- #1
chwala
Gold Member
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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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