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need to simplify this. $$\tan20^{\circ}\cos50^{\circ}+\cos40^{\circ}.$$

I have tried to express $\cos40^{\circ}$ in terms of $\sin20^{\circ}$ and $\cos 20^{\circ}$ but that does not help.

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how did you express cos 40 in terms of sin 20 and cos 20 –  EpicGuy Mar 21 at 10:32
    
@EpicGuy sounds like $cos(2\theta)$ –  Sabyasachi Mar 21 at 10:33
    
cos40=(cos20)^2-(sin20)^2 –  Anahit Mar 21 at 10:34
    
yes @Sabyasachi –  Anahit Mar 21 at 10:34

3 Answers 3

up vote 2 down vote accepted

Even easier: start with $\cos50^\circ=\sin40^\circ$ as suggested by Sabayasachi, then $$\tan20^\circ\cos50^\circ+\cos40^\circ=\frac{\sin20^\circ}{\cos20^\circ}2\sin20^\circ\cos20^\circ+1-2\sin^220^\circ$$ etc.

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Hint:Write $\cos(50^{\circ})$ as $\sin(40^{\circ})$ and $\tan(20^{\circ})$ as $\tan\left(\frac{40^{\circ}}2\right)$ using the formula:

$$\tan\left(\frac{x}{2}\right)=\frac{\sin(x)}{1+\cos(x)}$$

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$$\tan x\cos(90^\circ-2x)+\cos2x=\tan x\sin2x+\cos2x$$

Using Double angle formula,

this becomes $$t\cdot\frac{2t}{1+t^2}+\frac{1-t^2}{1+t^2}$$ where $t=\tan x$

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