Question

4sin(420-a)cos(60+a)

Answer 1

hi mate


your answer is in attachment

Answer 2

Answer:

$4\sin(420-a)\cos(60+a)=\sqrt3-2\sin 2a$

Step-by-step explanation:

Given : Expression $4\sin(420-a)\cos(60+a)$

To find : Solve the expression ?

Solution :

We can write the expression as

$=4\sin(360+(60-a))\cos(60+a)$

We know, $\sin (360+\theta)=\sin \theta$

$=4\sin(60-a)\cos(60+a)$

Applying formula,

$\sin (a-b)=\sin a \cos b-\cos a \sin b\\\cos (a+b)=\cos a \cos b-\sin a \sin b$

$=4(\sin 60 \cos a-\cos 60 \sin a)(\cos 60 \cos a-\sin 60 \sin a)$

$=4(\frac{\sqrt3}{2}\cos a-\frac{1}{2}\sin a)(\frac{1}{2} \cos a-\frac{\sqrt3}{2}\sin a)$

$=\frac{4}{4}[(\sqrt3\cos a-\sin a)(\cos a-\sqrt3\sin a)]$

$=\sqrt3\cos^2 a-\sin a\cos a-3\sin a\cos a+\sqrt3\sin^2a$

$=\sqrt3(\cos^2a+\sin^2a)-4\sin a\cos a$

$=\sqrt3(1)-4\sin a\cos a$

$=\sqrt3-2\sin 2a$

Therefore, $4\sin(420-a)\cos(60+a)=\sqrt3-2\sin 2a$