I have no idea how to prove this. Does anyone know where to start? We're allowed to use other trigonometric identities but i'm not sure why these are useful.
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.
27884
Problem that I am facing:  So naturally I decided to shift to the polar co-ordinates, however when I try to simplify the expression remains very complicated and despite using the fact that sin and cos belong to [-1,1] I cannot get any conclusive bound.  Thanks
86145
To prove:  $$\begin{align} \sin({\arccos{x}})&amp;=\sqrt{1-x^2}\\ \cos{\arcsin{x}}&amp;=\sqrt{1-x^2}\\ \sin{\arctan{x}}&amp;=\frac{x}{\sqrt{1+x^2}}\\ \cos{\arctan{x}}&amp;=\frac{1}{\sqrt{1+x^2}}\\ \tan{\arcsin{x}}&amp;=\frac{x}{\sqrt{1-x^2}}\\ \tan{\arccos{x}}&amp;=\frac{\sqrt{1-x^2}}{x}\\ \cot{\arcsin{x}}&amp;=\frac{\sqrt{1-x^2}}{x}\\ \cot{\arccos{x}}&amp;=\frac{x}{\sqrt{1-x^2}} \end{align}$$
27829
I've been messing around with modular arithmetic recently, and stumbled across this, but couldn't find a proof for it anywhere. I hate taking things as truth without knowing why, so could anyone provide a (fairly simple) proof?
105051
Obviously, there are some points (e.g. $\pi$, $30^\circ$) but I am unsure if there are more. How can it be proved that there are no more points, or, if there are, what those points will be?  EDIT: I largely meant to ask what is supposed in the first comment. Put another way, are there numbers such that $\sin x$ and $\arcsin x$ are both rational?
57930
Good morning! I'm having trouble with this problem... It's just taking me forever and I'm worn out and I'm lost on how to use a double angle identity for $72=2⋅36$  The problem reads as follows  An exact value for $\cos36°$ can be found using the following procedure. Begin by considering $\sin108°$. Note that $108=72+36$ and use the sine sum identity. Also note that $72=2⋅36$ and use double angle identities. If there are any common factors in each term, factor them out and cancel them if they are not equal to zero. You should eventually obtain a quadratic equation containing $\cos36°$. Use the quadratic formula to obtain the exact value for $\cos36°$. (Note that the quadratic formula should give two solutions. One can be disregarded - why?  Thank you in advance to anyone who can help.
38675
If $$\tan(\frac{\pi}{4}+\frac{\theta}{2})=\tan^3(\frac{\pi}{4}+\frac{\alpha}{2})$$, then prove that $$\sin \theta=\frac{3 \sin \alpha+\sin^3 \alpha}{1+3\sin^2 \alpha}$$  I tried using the fact that  $\frac{\cos A}{1-\sin A}=\tan(\frac{\pi}{4}+\frac{A}{2})$ but now not able to eliminate $\cos \alpha$ and $\cos \theta$. How should I proceed?
87602
Let     $$ 1&lt; \alpha_1&lt;\alpha_2&lt;......&lt;\alpha_n&lt; \pi/2$$ then prove that $$ tan(\alpha_1)&lt; \frac{sin(\alpha_1)+sin(\alpha_2)+.....+sin(\alpha_n)}{cos(\alpha_1)+cos(\alpha_2)+......+cos(\alpha_n)} &lt; tan(\alpha_n)$$   I tried to solve the problem but I am not actually able to get a well established answer or you can say I am not even able to advance ahead to solve this one. It tried to use AM-GM inequality, but its didn't work as I couldn't establish any direct relation.
102553
$$\tan\left(\frac{\pi}{2} -\theta \right) =  \cot \theta$$  I can prove this by changing into $\cos$ and $\sin$  But I want to know if it is possible to prove it using relation given below. If not why ? $$\tan\left(\frac{\pi}{2} -\theta \right) =\frac{ \tan\frac{\pi}{2} -\tan\theta}{1+ \tan\frac{\pi}{2}\tan\theta} $$
10869
No matter how I do it, I always end up with $x = 0, 90, 270$ and $360$. All of those except $270$ is right, but I can't quite figure out how to get the $270$ degrees out of the answer. I've tried using trig identities, I've tried squaring both sides, but I always end up with $$2\sin x\cos x$$ which then leads me to $x = 0, 90, 270, 360$. But $$\sin (270) + \cos (270) = -1$$ so I'm doing something wrong.
40295
I'd like to see a proof of this identity I discovered:  $$ \sum_{i=1}^{n}\sum_{j=1}^{i}a_j = \sum_{i=1}^{n}a_i(n-i+1) $$  Probably there's something I forgot because I couldn't manage it.  [edit]Possibly with just identities, not examples please!  [edit2]No induction procedure, either.
65202
Prove that $\tan20^°\tan40^°\tan60^°\tan80^°=3$    \begin{align} \tan20^°\tan40^°\tan60^°\tan80^°&amp;=\frac{\sin20^°\sin40^°\sin60^°\sin80^°}{\cos20^°\cos40^°\cos60^°\cos80^°} \\ &amp;=\frac{2^4(\sin20^°\sin40^°\sin60^°\sin80^°)^2}{\sin 160^°} \end{align}  I am stuck here.
51698
I most humbly beseech help for this question.  If $A+B+C=180$ degrees, then prove $$ \sin^2(A)+\sin^2(B)-\sin^2(C)=2\sin(A)\sin(B) \cos(C) $$ I am not sure what trig identity I should use to begin this problem.
39229
Verify that: $\arcsin \theta+\arccos \theta=\frac{\pi}{2}.$ (1)   How can one verify (1) when it is not generally true? We can rewrite (1) as      Verify that if $\sin u = \cos v$, then $u+v=\frac{\pi}{2}$.   What about $\sin \frac{2\pi}{3}$ and $\cos \frac{\pi}{6}$? $\sin \frac{2\pi}{3} = \cos \frac{\pi}{6}$, but $\frac{2\pi}{3}+\frac{\pi}{6} \neq \frac{\pi}{2}$.We may try to verify (1) as:  Let $\arcsin \theta=u$ and $\arccos \theta=v$, then $\sin u=\cos v = \theta.$ We can write   $$\sin(u+v)=\sin u \cos v+ \cos u \sin v,$$  which becomes   $$\sin(u+v)=\theta^2+ \cos u \sin v.$$  Now, $\cos u=\pm \sqrt{1-\theta^2}$, and $\sin v=\pm \sqrt{1-\theta^2}$. For $\cos u= \sqrt{1-\theta^2}$ and $\sin v= \sqrt{1-\theta^2}$, we have  $$\sin(u+v)= \theta^2+(1-\theta^2)=1,$$  $$\therefore u+v=\arcsin \theta + \arccos \theta = \arcsin(1)=\frac{\pi}{2}.$$ But $\arcsin(1)$ is also equal to $\frac{\pi}{2}+2n\pi, n \in \mathbb{Z}.$ What about that? And what about the cases when $\cos u = -\sqrt{1-\theta^2}$ and $\sin u = \sqrt{1-\theta^2}$ and the other way? Then $\sin(u+v)=2\theta^2-1=1\ \mbox{iff}\ \theta =\pm 1.$   I don't know where I am making a mistake, what assumptions are wrong, or where my reasoning goes wrong. Please help.
22236
$$\cos A \cdot \cos 2A  \cdot \cos 2^{2}A  \cdot \cos 2^{3}A  ... \cos 2^{n-1}A = \frac { \sin 2^n A}{ 2^n \sin  A } $$  I am very much inquisitive to see how this trigonometrical identity can be proved.  PS:I am not much of interested about an inductive proof.
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.
54211
I need to show the steps to prove this identity: $$\sin^4{x} = \frac{3-4\cos{2x}+\cos{4x}}{8}$$  I know that $\cos{2x}=\cos^2{x}-\sin^2{x}$. From there I do not know what to do. The solution should look like:  $$\sin^4{x}=sin^4{x}$$  I need to prove the right side equals the left side.