Advertisements
Advertisements
प्रश्न
Prove that sin(45° + θ) – sin(45° – θ) = `sqrt(2) sin θ`
Advertisements
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`
APPEARS IN
संबंधित प्रश्न
Find the values of `tan ((19pi)/3)`
If sin x = `15/17` and cos y = `12/13, 0 < x < pi/2, 0 < y < pi/2` find the value of sin(x + y)
Find the value of cos 105°.
Find the value of sin105°.
Find the value of tan `(7pi)/12`
Prove that sin 75° – sin 15° = cos 105° + cos 15°
Show that tan(45° + A) = `(1 + tan"A")/(1 - tan"A")`
If tan x = `"n"/("n" + 1)` and tan y = `1/(2"n" + 1)`, find tan(x + y)
Find the value of cos 2A, A lies in the first quadrant, when sin A = `4/5`
Prove that (1 + sec 2θ)(1 + sec 4θ) ... (1 + sec 2nθ) = tan 2nθ
Express the following as a sum or difference
sin 4x cos 2x
Show that `cos pi/15 cos (2pi)/15 cos (3pi)/15 cos (4pi)/15 cos (5pi)/15 cos (6pi)/15 cos (7pi)/15 = 1/128`
Prove that sin x + sin 2x + sin 3x = sin 2x (1 + 2 cos x)
Prove that cos(30° – A) cos(30° + A) + cos(45° – A) cos(45° + A) = `cos 2"A" + 1/4`
Show that cot(A + 15°) – tan(A – 15°) = `(4cos2"A")/(1 + 2 sin2"A")`
If A + B + C = `pi/2`, prove the following sin 2A + sin 2B + sin 2C = 4 cos A cos B cos C
If ∆ABC is a right triangle and if ∠A = `pi/2` then prove that sin2 B + sin2 C = 1
Choose the correct alternative:
Let fk(x) = `1/"k" [sin^"k" x + cos^"k" x]` where x ∈ R and k ≥ 1. Then f4(x) − f6(x) =
