I can't figure out how (cosx+isinx)^4 expands to 1-8 cos^2(x)+8 cos^4(x)-4 i cos(x) sin(x)+8 i cos^3(x) sin(x)  I got it equal to sin^4(x)+cos^4(x)+i (4 sin(x) cos^3(x)-4 sin^3(x) cos(x))-6 sin^2(x) cos^2(x)  How do I make it equal to 1-8 cos^2(x)+8 cos^4(x)-4 i cos(x) sin(x)+8 i cos^3(x) sin(x)
124176
Obtain another expression for $(\cos θ + i \sin θ)^4$ by direct multiplication (i.e., expand the bracket). Use the two expressions to show  $$ \cos 4\theta = 8 \cos^4 \theta − 8 \cos^2 \theta + 1,\\ \sin 4\theta = 8\cos^3\theta \sin\theta − 4 \cos \theta\sin \theta. $$ You may use the well-known identity: $\sin^2 \theta + \cos^2 \theta = 1$, but do not use any multiple angle formula.  I got using DMT that $(\cos \theta + i\sin \theta)^4 = (cos 4\theta + i\sin 4\theta)$ and using direct multiplication, I got $\cos^4\theta + \sin^4\theta - 4\cos^3\theta*\sin\theta*i - 4\cos\theta \sin^3\theta i - 6\cos^2\theta\sin^2\theta$
44532
I have to simplify and evaluate this :   $$\cot 70^\circ+4\cos 70^\circ$$  On evaluating it, the answer comes out to be $1.732$, or $\sqrt 3$ .  I tried to get everything in $\sin$ and $\cos$, but it doesn't go any further. Any hints?
42678
Let $C$ be the coefficient of $x^2$ in the expansion of the product $(1 - x)(1 + 2x)(1 - 3x)\cdots(1 + 14x)(1 - 15x).$ Find $|C|.$   Just to begin,  $(1-x)(1+2x) = -2x^2 + x + 1$  $(1-x)(1+2x)(1-3x) = 6x^3 - 5x^2 - 2x + 1$   But expanding on like this take too long.  In the end the terms will be like:  $(1 - 15x)(a_0 + a_1x^1 + a_2x^2 + ....)$  Then just looking at the $x^2$, it will be:  $(a_2x^2 - 15a_1x^2) = x^2(a_2 - 15a_1)$  But it it still a very hard problem, any hints?
81826
$(x+2)(x-1$)  Totally forgot how to do it. Isn't this the rule:  $(a+b)(c+d)$  $a*c +a*d +b*c +b*d$  The answer I'm getting is $x^2-x+2x-2$ which is wrong. Please help me out, thanks!
13485
I'm stuck with this integral  $\int\frac {\cos^4x}{\sin^3x} dx$    which I rewrote as  $\int \csc^3x \cos^4xdx$  then after using the half angle formula twice for $\cos^4x$ I got this  $\frac 14\int \csc^3x (1+\cos(2x))(1+\cos(2x))dx$   then after solving those products I got these integrals  $\frac 14 \{\int \csc^3xdx+2\int \csc^3x \cos(2x)dx + \int \csc^3x \cos^2(2x)dx\}$  I do know how to solve the $\int \csc^3xdx$ one but I'm totally lost on the other ones, any tips/help/advice would be highly appreciated! thanks in advance guys!
64473
$$\tan A + \tan 2A + \tan 3A=0$$ I tried converting these all in sin and cos and I got the answer but the answer didn't match.  So any one could just help me.  Thanks
38715
Evaluate $$\lim_{x\to \infty} \cos (\sqrt {x+1})-\cos (\sqrt x)$$   For this question I have tried using the squeeze theorem and some trigonometric manipulations such as using $$2\sin A\sin B=\cos (A-B) -\cos (A+B)$$.  But they were of no use. I also tried using the substitutions like $$x=1/y$$ Hence as $x\to \infty$ , $y\to 0$ but still no success.  I could've also tried using L'Hospital rule but I am not able to convert the function in appropriate indeterminate form. Any help would be greatly appreciated
77232
I need to use the fact that  $\tan 2x=\sin2x \ /\cos2x$ to prove that: $$\tan 2x=\frac{2\tan x}{1-\tan^2x}$$ I don't know where to start. Please help or hint. Thanks in advance.
82709
What I attempted thus far:  Multiplying by conjugate   $$\lim_{x \to 0} \frac{\sqrt{1 + \tan x} - \sqrt{1 + \sin x}}{x^3} \cdot \frac{\sqrt{1 + \tan x} + \sqrt{1 + \sin x}}{\sqrt{1 + \tan x} + \sqrt{1 + \sin x}} = \lim_{x \to 0} \frac{\tan x - \sin x}{x^3 \cdot (\sqrt{1 + \tan x} + \sqrt{1 + \sin x})}$$  factor out $\sin x$ in the numerator  $$\lim_{x \to 0} \frac{\sin x \cdot (\sec x - 1)}{x^3 \cdot (\sqrt{1 + \tan x} + \sqrt{1 + \sin x})}$$  simplify using $\lim_{x \to 0} \frac{\sin x}{x} = 1 $  $$\lim_{x \to 0} \frac{\sec x - 1}{x^2 \cdot (\sqrt{1 + \tan x} + \sqrt{1 + \sin x})}$$  From here I don't see any useful direction to go in, if I even went in an useful direction in the first place, I don't know.    I suspect that this could be evaluated using the definition of derivatives, if so, or not, any suggestions?
18581
I am a little weak in trigonometry. I have two questions:   Find the value of $\cos 52^{\circ} + \cos 68^{\circ} + \cos 172^{\circ} $ Find the value of $\sin 28^{\circ}+ \cos 17^{\circ} + \cos 28^{\circ} + \sin 17^{\circ} $   I am asking these questions because: 1. I am weak and unable to solve these. 2. I want to know the difference in these questions.
86145
To prove:  $$\begin{align} \sin({\arccos{x}})&amp;=\sqrt{1-x^2}\\ \cos{\arcsin{x}}&amp;=\sqrt{1-x^2}\\ \sin{\arctan{x}}&amp;=\frac{x}{\sqrt{1+x^2}}\\ \cos{\arctan{x}}&amp;=\frac{1}{\sqrt{1+x^2}}\\ \tan{\arcsin{x}}&amp;=\frac{x}{\sqrt{1-x^2}}\\ \tan{\arccos{x}}&amp;=\frac{\sqrt{1-x^2}}{x}\\ \cot{\arcsin{x}}&amp;=\frac{\sqrt{1-x^2}}{x}\\ \cot{\arccos{x}}&amp;=\frac{x}{\sqrt{1-x^2}} \end{align}$$
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!
5770
I have a dilemma. I have a task where I'm supposed to show that $|\cos z|^2+|\sin z|^2=1$ if and only if $z\in \mathbb{R} $. In my argument I have that $$\cos z\cdot\overline{\cos z} + \sin z\cdot\overline{\sin z}=\cos z\cdot \cos\overline{z} + \sin z\cdot \sin\overline{z}$$ but my teacher said that $\overline {\cos z} \neq \cos \overline{z}$.  Is this true?  But $$\overline{\cos(z)} = \overline{\frac{e^{iz}+e^{-iz}}{2}}=\frac{e^{-i\overline{z}}+e^{i\overline{z}}}{2}=\cos(\overline{z})$$
82310
To prove  $ \tan(A) + 2 \tan(2A) +4\tan4A + 8 \cot8A =\cot(A) $.  I tried to convert $\tan(4A)$ and $ \tan(8A)$ to $\tan(A$) and tried putting in L.H.S but it becomes a mess. Is there a easy and also intuitive  way to do this?  Thanks
72973
I have to prove the following statement.     $$1+\cos{2\pi\over5}+\cos{4\pi\over5}+\cos{6\pi\over5}+\cos{8\pi\over5}=0$$   I have tried to use the sum of angles formula for cosine, but didn't get to a point where I'd be able to show that it is equal to $0$.
67968
How is it possible to prove that the multiplication from 0 to n of $$\cos{(x/2^k)}$$ with $x=\pi / 4$ is $$\dfrac{\sqrt{2}}{2} \times \dfrac{\sqrt{2+\sqrt{2}}}{2} \times \dfrac{\sqrt{2+\sqrt{2 + \sqrt{2}}}}{2} \times  ...$$  I tried recurrence but got stuck
10042
Assume we have:  $$ \int (\cos{x} + \sin{x}\cos{x}) \, dx$$  Two ways to do it:  Use $$\sin{x}\cos{x} = \frac{ \sin{2x} }{2} $$ Then $$ \int \left(\cos{x} + \frac{\sin{2x}}{2} \right) \, dx = \sin{x} - \frac{ \cos{2x} }{ 4 } + C. $$ The other way, just see that $ u = \sin(x), du = \cos(x)dx $, $$ \int ( \cos{x} + \sin{x}\cos{x}) \, dx = \sin{x} + \frac{\sin^2{x}}{2} + C. $$  Now the part I don't see fully is, why aren't these results completely equal?  Taking the 2nd result, \begin{align} \sin{x} + \frac{\sin^2{x}}{2} &amp;= \sin{x} + \frac{1}{2} \, \left( \frac{1 - \cos{2x}}{2} \right) \\ &amp;= \sin{x} + \frac{1}{4} - \frac{\cos{2x}}{4}. \end{align}  So you have to absorb $\frac{1}{4}$ into $C$ for them to be equal.  Shouldn't they be equal straight away?