1. ## general solution

Gievn that tan A =3 and sin (A-B)=2 cos (A+B) , find tan B

i got tan B=-1/5

Find general solution , in rad cos b + cos 3b +cos 5b =0

cos b + 2 cos 4b cos b =0

cos b (1+2cos 4b )=0

cos b =0

and cos b = -1/2

I can only reach this far , i am not sure about the general solution >

Deduce $\cos^2 b+\cos^2 3b+\cos^2 5b=\frac{3}{2}$

no idea bout this .

2. Hello, thereddevils!

$\text{Given: }\:\tan A = 3\:\text{ and }\:\sin(A-B)\:=\:2\cos (A+B),\;\text{ find }\tan B$

We are given: . $\sin(A - B) \;=\;2\cos(A + B)$

Then: . $\sin A\cos B - \sin B\cos A \;=\;2(\cos A\cos B - \sin A\sin B)$

. . . . . $\sin A\cos B - \sin B\cos A \;=\;2\cos A\cos B - \sin A\sin B$

Divide by $\cos A\!:\quad \frac{\sin A\cos B}{\cos A} - \frac{\sin B\cos A}{\cos A} \;=\;\frac{2\cos A\cos B}{\cos A} - \frac{\sin A\sin B}{\cos A}$

. . . . . . . . . . . . . . . . $\underbrace{\tan A}_3\cos B - \sin B \;=\;2\cos B - 2\underbrace{\tan A}_3\sin B$

And we have: . $3\cos B - \sin B \;=\;2\cos B - 6\sin B$

. . . . . . . . . . . . . . . $5\sin B \;=\;-\cos B$

. . . . . . . . . . . . . . . . $\frac{\sin B}{\cos B} \;=\;-\frac{1}{5}$

. . . . . . . . . . . . . . . . $\boxed{\tan B \;=\;-\frac{1}{5}}$

3. Hello again, thereddevils!

Find the general solution in radians: . $\cos x + \cos3x +\cos5x \:=\:0$

$\cos x + 2\cos4x\cos x \:=\:0$

$\cos x (1+2\cos4x) \:=\:0$ . . . . Correct!

We have: . $\cos x \:=\:0 \quad\Rightarrow\quad \boxed{ x \:=\:\frac{\pi}{2} + \pi n}$

And: . $1+2\cos4x\:=\:0 \quad\Rightarrow\quad \cos4x \:=\:-\tfrac{1}{2}$

. . . $4x \:=\:\pm\frac{2\pi}{3} + 2\pi n \quad\Rightarrow\quad\boxed{ x \:=\:\pm\frac{\pi}{6} + \frac{\pi}{2}n }$

4. Hello thereddevils
Originally Posted by thereddevils
Deduce $\cos^2 b+\cos^2 3b+\cos^2 5b=\frac{3}{2}$
I think we have a problem here if this result is supposed to apply to all the solutions of the equation $\cos b+\cos 3b+\cos 5b=0$.

Look at the solution $b = \frac{\pi}{2}+n\pi = (2n+1)\frac{\pi}{2}$. In other words, $b$ is any odd multiple of $\frac{\pi}{2}$.

If $b$ is an odd multiple of $\frac{\pi}{2}$, then so are $3b$ and $5b$, and at these values $\cos b = \cos 3b = \cos 5b = 0$.

So, obviously, $\cos^2 b+\cos^2 3b+\cos^2 5b=0$ also.

However, when we consider the other set of values of $b$ (those for which $\cos 4b = -\tfrac12$) we do get the required result. Here's why:

$\cos^2 b+\cos^2 3b+\cos^2 5b=\tfrac12(1+\cos2b)+\tfrac12(1+\cos6b)+\tfrac12( 1+\cos10b)$

$=\tfrac32+\tfrac12(\cos2b+\cos6b+\cos10b)$

$=\tfrac32+\tfrac12(2\cos6b\cos4b+\cos6b)$

$=\tfrac32+\tfrac12\cos6b(2\cos4b+1)$

$=\tfrac32$, when $\cos4b = -\tfrac12$

So: are you sure you posted the complete question?