Consider the expression $a\sin\theta+b\cos\theta$. This expression consists of a combination of sine and cosine trigonometric functions which makes analysing it difficult. How can this expression be converted to an expression consisting of only sine or cosine functions?
13679
I don't understand this. These identities are given in the online notes for MIT's 18.01 calculus class. It's related to taking the sum of two trig functions and transforming them into a single trig function. I can use the formulas to do this, but I am having trouble finding anything on the internet to use a proof for my understanding.  Thanks for any responses!  EDITED (added after the first answer):  Sorry, the identity goes like this:  $A\sin{k(x-c)}=a\sin{kx}+b\cos{kx}$ ,  with the relationships  $a=A\cos{kc}$ , $b=-A\sin{kc}$ , $A=\sqrt{a^2+b^2}$ , and $\tan{kc}=-b/a$
38104
Question: If $\alpha$ and $\beta$ are the solutions of $a\cos \theta + b\sin \theta = c$, then show that: $$\cos (\alpha + \beta) = \frac{a^2 - b^2}{a^2 + b^2}$$  No idea how to even approach the problem. I tried taking two equations, by substituting $\alpha$ and $\beta$ in place of $\theta$ in the equation and manipulating them, but that didn't get me anywhere. Please help!
116573
Suppose that you have to prove the trig identity:  $$\frac{\sin\theta - \sin^3\theta}{\cos^2\theta}=\sin\theta$$  I have always been told that I should manipulate the left and right sides of the equation separately, until I have transformed them each into something identical. So I would do:  $$\frac{\sin\theta - \sin^3\theta}{\cos^2\theta}$$ $$=\frac{\sin\theta(1 - \sin^2\theta)}{\cos^2\theta}$$ $$=\frac{\sin\theta(\cos^2\theta)}{\cos^2\theta}$$ $$=\sin\theta$$  And then, since the left side equals the right side, I have proved the identity. My problem is: why can't I manipulate the entire equation? In this situation it probably won't make things any easier, but for certain identities, I can see ways to "prove" the identity by manipulating the entire equation, but cannot prove it by keeping both sides isolated.  I understand, of course, that I can't simply assume the identity is true. If I assume a false statement, and then derive from it a true statement, I still haven't proved the original statement. However, why can't I do this:  $$\frac{\sin\theta - \sin^3\theta}{\cos^2\theta}\not=\sin\theta$$ $$\sin\theta - \sin^3\theta\not=(\sin\theta)(\cos^2\theta)$$ $$\sin\theta(1 - \sin^2\theta)\not=(\sin\theta)(\cos^2\theta)$$ $$(\sin\theta)(\cos^2\theta)\not=(\sin\theta)(\cos^2\theta)$$  Since the last statement is obviously false, is this not a proof by contradiction that the first statement is false, and thus the identity is true?  Or, why can't I take the identity equation, manipulate it, arrive at $(\sin\theta)(\cos^2\theta)=(\sin\theta)(\cos^2\theta)$, and then work backwards to arrive at the trig identity. Now, I start with a statement which is obviously true, and derive another statement (the identity) which must also be true - isn't that correct?  Another argument that I have heard for keeping the two sides isolated is that manipulating an equation allows you to do things that are not always valid in every case. But the same is true when manipulating just one side of the equation. In my first proof, the step  $$\frac{\sin\theta(\cos^2\theta)}{\cos^2\theta}$$ $$=\sin\theta$$  is not valid when theta is $\pi/2$, for example, because then it constitutes division by zero.
79654
Question: Find the sum of the series:   $$\cos^3 \alpha +\cos^3 {3\alpha} + \cos^3 {5\alpha}+....+\cos^3 {(2n-1)\alpha}$$   The book from which this question was taken says that the answer is $\frac{3\sin{n\alpha}\cos{n\alpha}}{4\sin\alpha}+\frac{\sin{3n\alpha}\cos{3n\alpha}}{4\sin{3\alpha}}$.  My attempt to solve this question: $$\text{Let S be the trigonometric series,}$$ $$\cos {3\theta} = 4\cos^3\theta-3\cos\theta \implies 4\cos^3 \theta=\cos{3\theta}+3\cos\theta$$ Applying formula on $\cos^3\theta$,.. $$4S = 3\cos\alpha + \cos3\alpha +3\cos3\alpha + \cos9\alpha+...+3\cos{(2n-1)\alpha}+\cos{(6n-3)\alpha}$$ $$4S= 3(\cos \alpha + \cos 3\alpha+\cos5\alpha+...)+(\cos3\alpha + \cos9\alpha+..)$$ Applying the summation of cosine formula, $$4S= 3\frac{\sin{n\alpha}}{\sin\alpha}\cdot\cos{(\alpha+(n-1)\alpha)}+?$$  So my problem is I don't know how to apply the formula for the second series (denoted by '?') for a fixed number of terms $n$ or should I treat it as a general and separate series.  To clarify my doubt again, what I am asking is that would the $(2n-1)$ affect the formula for the second series (denoted by '?').  My working here looks most probably correct but if there is any mistake please correct it.
99574
How do I go about proving this? I know one method is:  $\eqalign{    \cos (90^\circ + \theta ) &amp;\equiv \cos90^\circ \cos\theta  - \sin90^\circ \sin\theta   \cr    &amp;  \equiv (0)(\cos\theta ) - (1)(\sin \theta )  \cr    &amp;  \equiv  - \sin \theta  \cr} $  I'd appreciate others, particularly ones that will allow me to visualize this identity on the unit circle.  Thanks alot.
22177
$$4\sin\theta +3\cos\theta  = 5$$   How would we solve this trigonometric equation using vectors? Since I'm not advanced, I do not truly know where to use vector product.   Regards!
118374
So in trig, say I have an acute angle $X$. And one can intuitively conclude that $\sin x + \cos x \ge 1$, but how does the fact that $$(\sin x + \cos x)^2 = 1 + 2 \sin x \cos x$$ tell me that it is true that $\sin x + \cos x \ge 1$? I don't quite see the connection. Thanks
79764
I would to prove that  $$1+\cos \theta+\cos 2\theta+\ldots+\cos n\theta =\displaystyle\frac{1}{2}+  \frac{\sin\left[(2n+1)\frac{\theta}{2}\right]}{2\sin\left(\frac{\theta}{2}\right)}$$ given that  $$1+z+z^2+z^3+\ldots+z^n=\displaystyle\frac {1-z^{n+1}}{1-z}$$ where $z\neq 1$.  I put $z=e^{i\theta}$. I already got in left hand side cos exp in real part, but there is a problem in the right hand side, I can't split imaginary part and real part. Please help me. Thanks in advance.
119123
Find the value of   $$1\sin 2^{\circ} +2\sin 4^{\circ} + 3\sin 6^{\circ}+\cdots+ 90\sin180^{\circ}$$   My attempt  I converted the $\sin$ functions which have arguments greater than $90^\circ$ to $\cos$ but I have gone no where with it!  I also tried using double angle formula for the angles which are even.
90876
If you take a linear combination of the cosine and sine function, then the result is , but with a new amplitude and phase shift. $$a \cos(\theta) + b \sin(\theta) = A \cos(\theta + \theta_0)$$ This is one of those facts which I've always felt should be more obvious to me than it is. I can't help feeling that I might be missing out on some good "conceptual reason" for this to be true. I did a bit of thinking about it today and I came up with the following explanation which I'm not particularly satisfied with. I feel like there should be some perspective on this which renders the fact obvious.      Multiplying the polar equation $r = 2a \cos(\theta) + 2b \sin(\theta)$ by $r$ yields $x^2+y^2 = 2a x + 2by$. Rearranging terms and adding $a^2 + b^2$ to both sides gives $x^2 - 2ax + a^2 +y^2 -2by + b^2 = a^2 + b^2$ which is identical to   $$(x-a)^2 + (y-b)^2 = a^2 + b^2$$   which we recognize as the equation for the circle with centre at $(a,b)$ and passing through the origin. This hints that $r = 2a \cos(\theta) + 2b \sin(\theta)$ should define a parametrization of this circle and, indeed, this holds. This circle can be rotated about the origin so as to have its centre on the $x$-axis. Such a circle will have the simpler equation   $$r = D \cos(\theta) $$   where $D=2(a^2+b^2)$ is the diameter of the circle. In polar coordinates, performing this rotation amounts to introducing a phase $\theta_0$. In fact $\theta_0$ is an argument for the point $(a,b)$.    Note I'm not really looking for rigorous proofs of this fact, more like interpretations, really.
89874
$$\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$$  I have attempted this question by expanding the left side using the cosine sum and difference formulas and then multiplying, and then simplifying till I replicated the identity on the right. I am not stuck. What's bothering me is that the way I went about this question seemed like a rather "clunky" method. I'm just curious if I've missed some underlying pattern that could have made it easier to reproduce the identity on the right.  The way I did it:  $$\begin{array}{l} \cos (A + B)\cos (A - B)\\ \equiv (\cos A\cos B - \sin A\sin B)(\cos A\cos B + \sin A\sin B)\\ \equiv {\cos ^2}A{\cos ^2}B - {\sin ^2}A{\sin ^2}B\\ \equiv {\cos ^2}A(1 - {\sin ^2}B) - (1 - {\cos ^2}A){\sin ^2}B\\ \equiv {\cos ^2}A - {\cos ^2}A{\sin ^2}B - {\sin ^2}B + {\cos ^2}A{\sin ^2}B\\ \equiv {\cos ^2}A - {\sin ^2}B\end{array}$$
56464
$ |A |= |B | = |C | = 1 $ ,where A B and C are complex nos  $$ \frac{A^2}{BC}+ \  \frac{B^2}{ \ {CA}} \ +\  \frac{C^2}{ \ {AB}} + 1 = 0$$  Find the possible values of $ |A + B + C |$  Tried substituting cos(theta) + i sin(theta)
95230
From the given figure, prove that      $$\cot\ \theta=\cot\ A+\cot\ B+\cot\ C$$     I could not even start to the first point. Please help me to solve this.  Thanks in advance
14641
I would like to confirm a solution. The question goes as:  Show that $$\frac{\sqrt{3}}{\sin20^{\circ}}-\frac{1}{\cos20^{\circ}}=4$$  Firstly I combined the two terms to form something like: $$\dfrac{\sqrt{3}\cos20^{\circ} - \sin20^{\circ}}{\sin20^{\circ}\cos20^{\circ}}.$$  Clearly, the numerator is of the form $a\cos\theta+b\sin\theta$, and can thus be expressed as $R\sin(\theta+\alpha)$, where $R=\sqrt{a^2+b^2}$, and $\alpha=\arctan(\dfrac{a}{b})$. Following this method, I obtain something like this: $$\dfrac{2\sin(20^{\circ}-60^{\circ})}{\sin20^{\circ}\cos20^{\circ}},$$ which results in the expression to be equal to $-4$. But, we took $R$ to be the principal square root, which was $2$. If we take it to be the negative root, $-2$, it follows that the expression is equal to $4$. So, can this expression have two values?   Edit: I would also like to mention how I computed the numerator:  Numerator:   $\sqrt{3}\cos20^{\circ} - \sin20^{\circ}$.  Now, this can be represented as $R\sin(\theta+\alpha)$. Computation of $R$: $R=\sqrt{a^2+b^2}\implies R=\sqrt{(3+1)}=2$ Computation of $\alpha$: $\tan\alpha=\dfrac{a}{b}\implies \tan\alpha=\dfrac{\sqrt3}{-1}\implies \alpha=-60^{\circ}$, because $\tan(-x)=-\tan x$, and in this case $\tan\alpha=-\sqrt3$, and $-\tan(60^{\circ})=-\sqrt3$, so $\alpha=-60^{\circ}$.  Hence, we conclude that numerator:$$2\sin(20+(-60))=2\sin(-40)=-2\sin(40^{\circ})$$
21098
Could someone help me with the steps for solving the below equation $$a \sin\theta + b \cos\theta = c$$ I know that the solution is $$\theta = \tan^{-1} \frac{c}{^+_-\sqrt{a^2 + b^2 - c^2}} - \tan^{-1} \frac{a}{b} $$ I just can't figure out the right steps to arrive at this solution.
1957
The proof of two angles for sine function is derived using $$\sin(A+B)=\sin A\cos B+\sin B\cos A$$ and $$\cos(A+B)=\cos A\cos B-\sin A\sin B$$ for cosine function. I know how to derive both of the proofs using acute angles which can be seen here  but pretty sure those who have taken trig know what I'm talking about. So I know how to derive and prove both of the two-angle functions using the acute angles, but what I am completely confused about is where those triangles came from. So for proving the two-angle cosine function, we look at two acute angles, $A$ and $B$, where $A+B&lt;90$ and keep on expanding. So my question is, where did those two triangles come from and what is the intuition behind having two acute triangles on top of each other?
117251
I came across this problem when I was at a book store inside of a book made to prepare Berkeley graduates to pass a mandatory exam. I wanted to buy the book, but, alas, I didn't have the money (forty bucks is a lot of money when you don't have a job). So I took my phone out and started taking as many pictures as I could. Unfortunately, I didn't take any pictures of the solutions!  Enough backstory. Time for math. The problem is as follows:  Prove that $\cos^p( \theta) \le \cos(p \theta)$ if $0\le\theta\le\frac\pi2$ and $0\le p\le 1$.  I tried using the series expansion for cosine, but that seemed to be a dead end. Then I tried using Euler's theorem, but I got stuck. Then I got distracted and started to think about other cosine identities. For example, $$\cos\left(\frac\theta2\right)=\pm\sqrt{\frac{1+\cos(\theta)}{2}}.$$ Then what is $\cos(\frac\theta3)$ equal to? I tried to figure it out then I realized that I needed to find the root of a cubic polynomial. Then I realized that I'm better at staying focused than finding the roots of cubic polynomials.  Anyways, a proof would be nice here. I really appreciate any hints or answers. I apologize for my digressions!