Question

Prove that:
cos 3A + cos 5A + cos 7A + cos 15A = 4 cos 4A cos 5A cos 6A​

Answer 1

Question :

Prove that

cos 3A + cos 5A + cos 7A + cos 15A = 4 cos 4A cos 5A cos 6A

Formula used :

$\sf\:1)\cos(a)+\cos(b)=2\cos(\dfrac{a+b}{2})\cos(\dfrac{a-b}{2})$

$\sf\:2)\cos(a)-\cos(b)=-2\sin(\dfrac{a+b}{2})\sin(\dfrac{a-b}{2})$

$\sf\:3)\sin(a)+\sin(b)=2\sin(\dfrac{a+b}{2})\sin(\dfrac{a-b}{2})$

$\sf\:4)\sin(a)-\sin(b)=2\cos(\dfrac{a+b}{2})\sin(\dfrac{a-b}{2})$

Solution :

We have to prove that

cos 3A + cos 5A + cos 7A + cos 15A = 4 cos 4A cos 5A cos 6A

$\sf\blue{LHS}=\cos3A+\cos5A+\cos7A+\cos15A$

$\sf=[\cos15A+\cos3A]+[\cos7A+\cos5A]$

Now use the formula of $\sf\green{\cos\:a+\cos\:b}$

then ,

$\sf=2\cos(\dfrac{15A+3A}{2})\cos(\dfrac{15A-3A}{2})+2\cos(\dfrac{7A+5A}{2})\cos(\dfrac{7A-5A}{2})$

$\sf=2\cos(\dfrac{18A}{2})\cos(\dfrac{12A}{2})+2\cos(\dfrac{12A}{2})\cos(\dfrac{2A}{2})$

$\sf=2\cos9A\cos6A+2\cos6A\cos\:A$

$\sf=2\cos6A[\cos9A+\cos\:A]$

Again use the formula of $\sf\pink{\cos\:a+\cos\:b}$

then ,

$\sf=2\cos6A[2\cos(\dfrac{9A+A}{2})\cos(\dfrac{9A-A}{2})]$

$\sf=2\cos6A[2\cos(\dfrac{10A}{2})\cos(\dfrac{8A}{2})]$

$\sf=2\cos6A[2\cos5A\cos4A]$

$\sf=4\cos6A\cos5A\cos4A$

$\sf\blue{RHS}=4\cos6A\cos5A\cos4A$

Now, $\sf\pink{LHS=RHS}$

Hence Proved

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More Formula's:

$\sf1)\sin(a+b)=\sin\:a\cos\:b+\sin\:b\cos\:a$

$\sf2)\sin(a-b)=\sin\:a\cos\:b-\sin\:b\cos\:a$

$\sf3)\cos(a+b)=\cos\:a\cos\:b-\sin\:b\sin\:a$

$\sf4)\cos(a-b)=\cos\:a\cos\:b+\sin\:b\sin\:a$