Advertisements
Advertisements
Question
If cosθ + sinθ = `sqrt2` cosθ, show that cosθ - sinθ = `sqrt2` sinθ.
Advertisements
Solution 1
(cosθ + sinθ)2 = (`sqrt2`. cosθ)2
cos2θ + sin2θ + 2.cosθ.sinθ = 2cos2θ
1 + 2.cosθ.sinθ = 2cos2θ
2.cosθ.sinθ = 2cos2θ − 1
(cosθ.sinθ)2 = cos2θ + sin2θ − 2.cosθ.sinθ
= 1 − (2.cos2θ − 1)
= 1 − 2.cos2θ +1
= 2 − 2.cos2θ
= 2(1 − cos2θ)
cosθ − sinθ = `sqrt(2sin^2θ)`
= `sqrt2`sinθ
Hence proved
Solution 2
RELATED QUESTIONS
Prove the following trigonometric identities.
`(tan^2 A)/(1 + tan^2 A) + (cot^2 A)/(1 + cot^2 A) = 1`
Prove that `(sec theta - 1)/(sec theta + 1) = ((sin theta)/(1 + cos theta))^2`
Prove the following identities:
`(sin theta + 1 - cos theta)/(cos theta - 1 + sin theta) = (1 + sin theta)/(cos theta)`
Find the value of sin ` 48° sec 42° + cos 48° cosec 42°`
If sec θ + tan θ = x, write the value of sec θ − tan θ in terms of x.
Prove the following identity:
tan2A − sin2A = tan2A · sin2A
Prove the following identities.
cot θ + tan θ = sec θ cosec θ
If x sin3 θ + y cos3 θ = sin θ cos θ and x sin θ = y cos θ, then prove that x2 + y2 = 1
If 2sin2β − cos2β = 2, then β is ______.
Prove that `(1 + tan^2 A)/(1 + cot^2 A)` = sec2 A – 1
