- #1

chwala

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- Homework Statement
- See attached.

- Relevant Equations
- Trigonometry

For part (a),

We know that ##\cos (-θ)=\cos (θ)## and ##\sin (-θ)=-\sin (θ)##

##\cos (A-B)=\cos A\cos (-B) -\sin A\sin(-B)##

##\cos (A-B)=\cos A\cos (B) +\sin A\sin(B)##

##\cos (A-B)=\cos A\cos B+\sin A\sin B##

For part (b)

...

##f(θ)=\cos 60^0- \sin (θ+30^0)\sin (θ-30^0)##

##f(θ)=\cos 2θ+ \sin (θ+30^0)\sin (θ-30^0)##

**on addition**,

...

##2f(θ)= \cos 60^0 + cos 2θ##

##2f(θ)=\dfrac{1}{2}+ (2\cos^2 θ -1)##

##f(θ)=\cos^2 θ - \dfrac{1}{4}##For part (c) i,

The maximum value of ##f(θ)## is given by,

##f(180^0)=\left[1-\dfrac{1}{4}\right]=\dfrac{3}{4}## at smallest positive value of ##θ=180^0##

The minimum value of ##f(θ)## is given by,

##f(90^0)=\left[0-\dfrac{1}{4}\right]=-\dfrac{1}{4}## at smallest positive value of ##θ=90^0##

Bingo!

Any insight/alternative approach is highly welcome. Cheers guys.

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