Problem: If $\frac{\sin^4\theta}{x}+\frac{\cos^4\theta}{y}=\frac{1}{x+y}$, show that, $\frac{\sin^{2m+2}\theta}{x^m}+\frac{\cos^{2m+2}\theta}{y^m}=\frac{1}{(x+y)^m}$.  My attempt: I observed that if both the sides of the given equation are raised to the power of m, the R.H.S of the resulting equation matches with the R.H.S of the result that has to be proved. After that we need to simplify the L.H.S of the resulting expression to match it with the L.H.S of the result that has to be proved.  $\frac{\sin^4\theta}{x}+\frac{\cos^4\theta}{y}=\frac{1}{x+y}$  $\implies(\frac{\sin^4\theta}{x}+\frac{\cos^4\theta}{y})^m=(\frac{1}{x+y})^m$  $\implies(\frac{y\sin^4\theta+x\cos^4\theta}{xy})^m=\frac{1}{(x+y)^m}$  $\implies\frac{(y\sin^4\theta+x\cos^4\theta)^m}{x^my^m}=\frac{1}{(x+y)^m}$  My problem: I am stuck at this step and cannot understand how to proceed, that is, how to simplify $(y\sin^4\theta+x\cos^4\theta)^m$. Please help. A continuation of my method as well as alternate methods (except mathematical induction) are welcome.
92277
If $\frac{\sin^4 x}{a}+\frac{\cos^4 x}{b}=\frac{1}{a+b}$, then show that $\frac{\sin^6 x}{a^2}+\frac{\cos^6 x}{b^2}=\frac{1}{(a+b)^2}$     My work: $(\frac{\sin^4 x}{a}+\frac{\cos^4 x}{b})=\frac{1}{a+b}$ By squaring both sides, we get, $\frac{\sin^8 x}{a^2}+\frac{\cos^8 x}{b^2}+2\frac{\sin^4 x \cos^4 x}{ab}=\frac{1}{(a+b)^2}$ $\frac{\sin^6 x}{a^2}+\frac{\cos^6 x}{b^2}-2\frac{\sin^4 x \cos^4 x}{ab}-\frac{\sin^6 x \cos^2 x}{a^2}-\frac{\sin^2 x \cos^6 x}{b^2}=\frac{1}{(a+b)^2}$ So, now, we have to prove that, $-2\frac{\sin^4 x \cos^4 x}{ab}-\frac{\sin^6 x \cos^2 x}{a^2}-\frac{\sin^2 x \cos^6 x}{b^2}=0$ I cannot do this. Please help!
20760
I used the following way but got wrong answer $$A.M. \ge  G.M.$$ $$ \frac{\sin \theta + \csc  \theta}{2} \ge \sqrt{\sin \theta \cdot \csc \theta}$$ Squaring both sides, \begin{equation*} (\sin\theta + \csc\theta )^2 \ge 4 \tag{1}  \end{equation*} Similarly \begin{equation*} (\cos\theta + \sec\theta )^2 \ge 4 \tag{2} \end{equation*} Adding equation (1)  and (2)  \begin{equation*} (\sin\theta + \csc\theta )^2+(\cos\theta + \sec\theta )^2 \ge 8 \end{equation*} What is wrong?
120874
Prove the following trigonometric identity: $$1 + \tan^2\theta = \sec^2\theta$$  I'm curious to know of the different ways of proving this depending on different characterizations of tangent and secant.
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.
45540
Express $\sqrt{3}\sin\theta - \cos\theta$ as: $a\cos (\theta + \alpha) $  Can someone please explain to me how to go about doing this?
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.
11231
Another problem that I already wasted hours on. Given $$4\sinθ +3\cosθ  = 5$$  Find   $$4\cosθ -3\sinθ$$  Help me guys (PS:I'm not that good in maths)
33692
If $x+y+z=xyz$, prove $\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$ given that $x^2~,~y^2~,~z^2\ne1$  I came across this question in an ancient ($19$th century) Trigonometry book, and this is the method they use to prove the result (please note: I understand this method fully): Let $x=\tan A$ ,$~y=\tan B$ and $z=\tan C$ which is acceptable without any loss of generality. This means that we are saying that $$\tan A+\tan B+\tan C=\tan A\tan B\tan C$$ Consider $\tan (A+B+C)$: $$\tan (A+B+C)=\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1- \tan A \tan B- \tan C \tan A - \tan B \tan C}$$ So if $~\tan A+\tan B+\tan C=\tan A\tan B\tan C~~$ then $~~\tan (A+B+C)=0$. Hence, let $~A+B+C=\pi$. Now consider $\tan (2A+2B+2C)$: $$\tan (2A+2B+2C)=\frac{\tan 2A+\tan 2B+\tan 2C-\tan 2A\tan 2B\tan 2C}{1- \tan 2A \tan 2B- \tan 2C \tan 2A - \tan 2B \tan 2C}=0$$ $$\implies \tan 2A+\tan 2B+\tan 2C=\tan 2A\tan 2B\tan 2C$$ $$\implies \frac{2\tan A}{1-\tan^2 A}+\frac{2\tan B}{1-\tan^2 B}+\frac{2\tan C}{1-\tan^2 C}=\frac{2\tan A}{1-\tan^2 A}\times\frac{2\tan B}{1-\tan^2 B}\times\frac{2\tan C}{1-\tan^2 C}$$ $$\therefore\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$$ as required.  My question is, is there any other way of proving this without this rather heavy use of trigonometry? I also would prefer not to work through heaps of algabraic manipulation and expansion to obtain the required result, although if that's necessary I will put up with it ;) Thank you for your help.
14883
How do you prove the following trigonometric identity: $$ \sin^2\theta+\cos^2\theta=1$$  I'm curious to know of the different ways of proving this depending on different characterizations of sine and cosine.
9253
The question states:    I'm not sure what to do here. I used the geometric sum to do $2^{-n}$, but I can't think of a way to sum $\sin(n\theta)$ to get the required RHS.    Further working     Apologies for maybe being slow or missing something obvious here... I don't know how to get it into required form.
37910
$m,n$ are integers bigger than 0.   I need to prove:  $$\frac{1}{\sqrt[n]{1+m}}+\frac{1}{\sqrt[m]{1+n}} \geq 1$$   I tried to multiply both sides by the common factor and raise both sides to the power of $m*n$ but it did not work for me, and I have no other idea how to proceed.  I don't want the full solution, but just a hint.  Thank you.
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
88959
Prove that $$\frac{x^2}{1+x^4} \leq \frac{1}{2}.$$   First off, we observe that the expression on the LHS is positive for all $x \in \Bbb R,$ and equality is achieved iff $x \in \{-1, 1 \}$. That being said, we start by manipulating the expression as below;  \begin{align} \frac{x^2}{1+x^4} - \frac{1}{2} &amp;\leq 0\\\\ \impliedby\frac{2x^2 - (1+x^4)}{2(1+x^4)} &amp;\leq 0\\\\ \impliedby -\frac{(x^4 - 2x^2 + 1)}{2(1+x^4)}&amp;\leq 0\\\\ \impliedby \frac{x^4 - 2x^2 + 1}{2(1+x^4)} &amp;\geq 0\\\\ \end{align}   We then make the observation that $x^4 - 2x^2 + 1 = (x^2 - 1)^2 \geq 0$ for all $x \in \Bbb R$ and that $1 + x^4 \geq 1$ for all $x \in \Bbb R$, and so the final inequality holds for all $x \in \Bbb R. \\\hspace{92pt}\square$  Is it okay to prove it in this way? I have no solutions so it would be nice if I could have some validation!
30830
$$ 0&lt;\theta&lt;\pi/2$$   and $$\sin\theta+\cos\theta+\tan\theta+\cot\theta+\sec\theta+\csc\theta=7$$   then show that $\sin 2\theta$ is a root of the equation $$x^2 -44x +36=0$$   I tried to use the above given equation of all the trigonometric ratios, though I ended up with an expression of $\sin\theta\cos\theta$.   But this too is in the form of $\sin\theta$ and $\cos\theta$ which was $\sin\theta\cos\theta=\frac{1}{6-\sin\theta-\cos\theta}$.  When I put that in the quadratic equation, it would again transform into the form of $\sin\theta$ and $\cos\theta$, and hence at last I couldn't prove the thing.  Even hints would work, as I would like to solve the question myself.
84743
$$\frac {\sin x}{1-\cos x} = \frac {1+\cos x}{\sin x}$$ The only way I can see of doing this is by cross multiplying but isn't that not allowed when trying to prove something?
41148
I'm doing some math exercises but I got stuck on this problem.  In the book of Bogoslavov Vene it says to prove that:  $$\frac{\sin x}{1-\cos x} = \frac{1+\cos x}{\sin x}.$$  It is easy if we do it like this: $\sin^2 x = (1-\cos x)(1+\cos x)=1-\cos^2 x$. But how to prove it just by going from the left part or from the right  Can anybody help me?  Thank you!
26665
The following problem was on a math competition that I participated in at my school about a month ago:      Prove that the equation $\cos(\sin x)=\sin(\cos x)$ has no real solutions.   I will outline my proof below.  I think it has some holes.  My approach to the problem was to say that the following equations must have real solution(s) if the above equation has solution(s):  $$ \cos^2(\sin x)=\sin^2(\cos x)\\ 1-\cos^2(\sin x)=1-\sin^2(\cos x)\\ \sin^2(\sin x)=\cos^2(\cos x)\\ \sin(\sin x)=\pm\cos(\cos x)\\ $$  I then proceeded to split into cases and use the identity $\cos t = \sin(\frac{\pi}{2} \pm t\pm y2\pi)$ to get   $$ \sin x=\frac{\pi}{2} \pm \cos x\pm y2\pi\\ $$  and the identity $-\cos t = \sin(-\frac{\pi}{2}\pm t\pm y2\pi) $ to get   $$ \sin x=- \frac{\pi}{2}\pm \cos x \pm y2\pi.\\ $$  where $y$ is any integer.  I argued that $y=0$ was the only value of $y$ that made any sense (since the values of sine and cosine remain between $-1$ and $1$).  Therefore, the above equations become   $$ \sin x=\frac{\pi}{2} \pm \cos x\implies \sin x \pm \cos x=\frac{\pi}{2}\\ $$  and   $$ \sin x=- \frac{\pi}{2}\pm \cos x\implies \cos x\pm\sin x= \frac{\pi}{2}.\\ $$  Then, by a short optimization argument, I showed that these last two equations have no real solutions.  First, does this proof make sense?  Second, if my proof makes sense, then I feel that it was not very elegant nor simple.  Is my approach the best, or is there a better (i.e., more elegant, shorter, simpler) proof?