Advertisements
Advertisements
प्रश्न
Prove that cos(A + B) cos(A – B) = cos2A – sin2B = cos2B – sin2A
Advertisements
उत्तर
L.H.S = cos(A + B) cos(A – B)
= (cos A cos B – sin A sin B)(cos A cos B + sin (A sin B)
= cos2A cos2B – sin2A sin2B
= cos2A (1 – sin2B) – (1 – cos2A) sin2B
= cos2A – cos2A sin2B – sin2B + cos2A sin2B
= cos2A – sin2B
= R.H.S
Now cos2A – sin2B = (1 – sin2A) – (1 – cos2B)
= 1 – sin2A – 1 + cos2B
= cos2B – sin2A
APPEARS IN
संबंधित प्रश्न
Find the value of the trigonometric functions for the following:
cos θ = `2/3`, θ lies in the I quadrant
Find the value of the trigonometric functions for the following:
tan θ = −2, θ lies in the II quadrant
If sin x = `15/17` and cos y = `12/13, 0 < x < pi/2, 0 < y < pi/2`, find the value of tan(x + y)
Find cos(x − y), given that cos x = `- 4/5` with `pi < x < (3pi)/2` and sin y = `- 24/25` with `pi < y < (3pi)/2`
Find the value of sin105°.
Prove that cos(π + θ) = − cos θ
Find a quadratic equation whose roots are sin 15° and cos 15°
Prove that sin(45° + θ) – sin(45° – θ) = `sqrt(2) sin θ`
Show that tan 75° + cot 75° = 4
If tan x = `"n"/("n" + 1)` and tan y = `1/(2"n" + 1)`, find tan(x + y)
If θ + Φ = α and tan θ = k tan Φ, then prove that sin(θ – Φ) = `("k" - 1)/("k" + 1)` sin α
If θ is an acute angle, then find `sin (pi/4 - theta/2)`, when sin θ = `1/25`
Show that `cot(7 1^circ/2) = sqrt(2) + sqrt(3) + sqrt(4) + sqrt(6)`
Express the following as a product
cos 65° + cos 15°
Prove that `(sin 4x + sin 2x)/(cos 4x + cos 2x)` = tan 3x
If A + B + C = 180°, prove that `tan "A"/2 tan "B"/2 + tan "B"/2 tan "C"/2 + tan "C"/2 tan "A"/2` = 1
If x + y + z = xyz, then prove that `(2x)/(1 - x^2) + (2y)/(1 - y^2) + (2z)/(1 - z^2) = (2x)/(1 - x^2) (2y)/(1 - y^2) (2z)/(1 - z^2)`
Choose the correct alternative:
If `pi < 2theta < (3pi)/2`, then `sqrt(2 + sqrt(2 + 2cos4theta)` equals to
Choose the correct alternative:
cos 1° + cos 2° + cos 3° + ... + cos 179° =
