Trigonometric Ratios and Identities

We have, f(x)=3sin‌12x+4cos10x
Period of 3sin‌12x is π.
Period of 4cos16x is π.
∴ Period of f(x)=3sin‌12x+4cos16x is π
Maximum of f(x)= Maximum of f(x) in [0,π]
f(x)=3sin‌12x+4cos16x
On differentiating w.r.t. x, we get
‌f′(x)=36sin‌11x‌cos‌x+64cos15x(−sin‌x)
‌⇒f′(x)=4sin‌x‌cos‌x(9sin‌10x−16cos14x)
‌⇒f′(x)=2sin‌2x(9sin‌10x−16cos14x)
On putting f′(x)=0, we get
‌2sin‌2x(9sin‌10x−16cos14x)=0
‌⇒‌‌2sin‌2x=0⇒x=0,‌

π

2

,π
‌‌ x ‌0r∕2r
‌f(x)43‌‌4
Clearly, the maximum value of f(x) is 4 .