Mathematics Stack Exchange
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Questions tagged [trigonometry]

Questions about trigonometric functions (both geometric and circular), relationships between lengths and angles in triangles and other topics relating to measuring triangles.

Trigonometry is a branch of mathematics that studies relationships involving lengths and angles of triangles.

Trigonometry is most simply associated with planar right-angle triangles. The applicability to non-right-angle triangles exists, but, since any non-right-angle triangle (on a flat plane) can be bisected to create two right-angle triangles, most problems can be reduced to calculations on right-angle triangles. Thus the majority of applications relate to right-angle triangles.

One exception to this is spherical trigonometry, the study of triangles on spheres, surfaces of constant positive curvature, in elliptic geometry. Trigonometry on surfaces of negative curvature is part of hyperbolic geometry.

Trigonometric identities are equalities that involve trigonometric functions and are true for every single value of the occurring variables. Geometrically, these are identities involving certain functions of one or more angles.

See Wikipedia's list of trigonometric identities.

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Verify the identity $\frac{\tan(a+b)}{\tan(a-b)}$ = $\frac{\sin(a)\cos(a)+\sin(b)\cos(b)}{\sin(a)\cos(a)-\sin(b)\cos(b)}$

I've been asked to verify the following identity but I don't know how to do it. $$\frac{\tan(a+b)}{\tan(a-b)} = \frac{\sin(a)\cos(a)+\sin(b)\cos(b)}{\sin(a)\cos(a)-\sin(b)\cos(b)}$$ When I try I get $$\frac{\tan(a+b)}{\tan(a-b)} =…
maybedave
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Establishing $\frac{ \sin mx}{\sin x}=(-4)^{(m-1)/2}\prod_{1\leq j\leq(m-1)/2}\left(\sin^2x-\sin^2\frac{2\pi j}{m}\right) $ for odd $m$

Let $m$ be an odd positive integer. Prove that $$ \dfrac{ \sin (mx) }{\sin x } = (-4)^{\frac{m-1}{2}} \prod_{1 \leq j \leq \frac{(m-1)}{2} } \left( \sin^2 x - \sin^2 \left( \dfrac{ 2 \pi j }{m } \right) \right) $$ Atempt to the proof My idea is…
James
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Prove that $ax = \cos(\pi\cdot x)$ has exactly one solution

Prove that $ax = \cos(\pi\cdot x)$ has exactly one solution when $0 \le x \le 1$. a is any positive real number. I can solve this question fine by drawing $\cos(\pi\cdot x)$ out but it's considered informal. I need help on a written proof.
meiryo
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If $\frac{\sin^4x}{a}+\frac{\cos^4x}{b}=\frac{1}{a+b},$ then show that $\frac{\sin^6x}{a^2}+\frac{\cos^6x}{b^2}=\frac{1}{(a+b)^2}$

Question: If $\frac{\sin^4x}{a}+\frac{\cos^4x}{b}=\frac{1}{a+b},$ then show that $\frac{\sin^6x}{a^2}+\frac{\cos^6x}{b^2}=\frac{1}{(a+b)^2}$. My approach: Since $$\frac{\sin^4x}{a}+\frac{\cos^4x}{b}=\frac{1}{a+b} \\ \implies…
Sanket Biswas
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What is the general solution to the equation $\sin x + \sqrt{3}\cos x = \sqrt2$

I need to find the general solution to the equation $$\sin(x) + \sqrt3\cos(x)=\sqrt2$$ So I went ahead and divided by $2$, thus getting the form $$\cos(x-\frac{\pi}{6})=\cos(\frac{\pi}{4})$$ Thus the general solution to this would be $$x = 2n\pi…
Techie5879
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How to calculate (by hand) trig functions?

I asked a similar question in high school and people started laughing as they thought like it's trivial. We only learn to memories the values cosine function but not even in college we're we taught how exactly it works. Trig functions as far as I…
Matko
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Evaluating $\cos(A+B)$, given $\cos A$ and $\sin(B)$

Find the exact value: Find $\cos(A+B)$ given that $\cos A=1/3$, with $A$ in the first quadrant, and $\sin B = -1/4$, with $B$ in the fourth quadrant.
flare
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Solving $\sin(2x) + 3\cos(2x) = 0$

Okay, there's this simple equation I've been looking into for a while and I don't know why one way of solving it is not correct. See: $$\sin(2x) + 3\cos(2x) = 0$$ Well, the most obvious would be to rearrange to get: $$\tan(2x)=-3$$ and get the…
Jerry
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How to determine the exact value of $\sin(585^\circ)$?

I'm clueless on this question. Could someone explain how to do it?
missiledragon
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Solve the equation $\min \{ \sin x, \cos x \} = \frac{\pi}{4}$ in $[0, 2\pi]$.

Consider the following equation: $$\min \{ \sin x, \cos x \} = \dfrac{\pi}{4}$$ I have to solve this equation for $x$ in $[0, 2\pi]$. What I want to know is how can I solve this without going to a site like Desmos and plotting $y = \min \{ \sin x,…
user592938
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Find $\arctan x_1 \cdot \arctan x_2$, where $x_1$ and $x_2$ are roots of $x^2 - 2\sqrt{2}x + 1 = 0$.

I am told that $x_1$ and $x_2$ are the roots of the following equation: $$ x^2 - 2\sqrt{2}x + 1 = 0 $$ And I have to find the following: $$\hspace{6cm} \arctan x_1 + \arctan x_2 \hspace{5cm} (1)$$ $$\hspace{6.cm} \arctan x_1 \cdot \arctan x_2…
user592938
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If sum of the series $\frac {\tan 1}{\cos 2}+\frac{\tan 2}{\cos 4} +\frac{\tan 3}{\cos 6}...\frac{\tan 1024}{\cos 2048}=\tan \lambda -\tan \mu$

Find the value of $\lambda +\mu$ The expression is $$\frac{\sin 1}{\cos 1 \cos 2}+\frac{\sin 2}{\cos 2 \cos 4}......$$ $$\frac {2\sin^21}{\sin 2\cos 2} +\frac{2\sin^22}{\sin 4\cos 4}$$ and so on, but that isn’t getting me anywhere. What can be…
Aditya
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Rewriting expression to use at most one trigonometric function

I have the expression: $$41\sqrt{2}\cos(v) + 41\sqrt{2}\sin(v)$$ And I want to rewrite it like an expression in v ∈ R that contains at most one trigonometric function. What I have tried to do is: $$41\sqrt{2}(\cos(v) + \sin(v))$$ But now I don't…
Blue shirt
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Trying to solve equation $\sin^2 x = 1$

The assignment is that I'm suppose to correct a flawed solution to the equation $\sin^2 x = 1$. The flawed solution is: $\sin^2 x = 1$ $\sin x = 1$ $x = 90^\circ + 2n\pi$ I thought I was simply to correct the fact that they forgot the negative…
Johan
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Can't find solutions for $\tan{2x} = \tan{x}$

Solving $\tan{2x}=\tan{x}$ Reducing the left side: $$\frac{\sin{2x}}{\cos{2x}} = \frac{2\sin{x}\cos{x}}{2\cos^{2}(x)-1}.$$ Reducing the right side: $$\tan{x} = \frac{\sin{x}}{\cos{x}}.$$ therefore: $$\frac{2\sin x\cos x}{2\cos^2x-1} = \frac{\sin…
hondaman
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