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प्रश्न
Prove that sin(45° + θ) – sin(45° – θ) = `sqrt(2) sin θ`
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उत्तर
sin (45° + θ) – sin (45° – θ) = `sqrt(2) sin θ`
sin(45° + θ) – sin(45° – θ) = (sin 45° cos θ + cos 45° sin θ) – (sin 45° cos θ + cos 45° sin θ)
= sin 45° cos θ + cos 45° sin θ – sin 45° cos θ + cos 45° sin θ
= 2 cos 45° sin θ
= `2 xx 1/sqrt(2) sin theta`
= `2/sqrt(2) xx sqrt(2)/sqrt(2) xx sin theta`
sin(45° + θ) – sin(45° – θ) = `(2sqrt(2))/2 sin theta`
= `sqrt(2) sin theta`
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