if root 3 tan theta = 3 sin theta, find the value of sin² theta - cos² theta
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3tantheta=9sin^2theta
sintheta/costheta=3sin^2theta
1/3=sinthetacostheta
1/3=2/2sinthetacostheta
1/3=1/2sin2theta
2/3=sin2theta ----------(1)
cos^2theta-sin^2theta=cos2theta
So
sin^2theta-cos^2theta= --cos2theta--------(2)
From eq 1
sin2theta=2/3
we know cos^2(2theta)= 1--sin^2(2theta)
cos^2(2theta)=1--4/9
cos^2(2theta)=5/9
cos(2theta)=root{(5)}/3
So from eq (2)
sin^2theta-cos^2theta= --cos2theta--------(2)
substituting:
==》
sintheta/costheta=3sin^2theta
1/3=sinthetacostheta
1/3=2/2sinthetacostheta
1/3=1/2sin2theta
2/3=sin2theta ----------(1)
cos^2theta-sin^2theta=cos2theta
So
sin^2theta-cos^2theta= --cos2theta--------(2)
From eq 1
sin2theta=2/3
we know cos^2(2theta)= 1--sin^2(2theta)
cos^2(2theta)=1--4/9
cos^2(2theta)=5/9
cos(2theta)=root{(5)}/3
So from eq (2)
sin^2theta-cos^2theta= --cos2theta--------(2)
substituting:
==》
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