Advertisements
Advertisements
प्रश्न
Prove that sin 105° + cos 105° = cos 45°
Advertisements
उत्तर
sin 105° + cos 105° = sin(90° + 15°) + cos(90° + 15°)
= cos 15° – sin 15°
= cos(45° – 30°) sin(45° – 30°)
= (cos 45° . cos30° + sin 45° sin 30°) – (sin 45° cos 30° – cos 45° sin 30°)
= `(1/sqrt(2) * sqrt(3)/2 + 1/sqrt(2) * 1/sqrt(2)) - (1/sqrt(2) * sqrt(3)/2 - 1/sqrt(2) * 1/sqrt(2))`
= `sqrt(3)/(2sqrt(2)) + 1/(2sqrt(2)) - sqrt(3)/(2sqrt(2)) + 1/(2sqrt(2))`
= `2/(2sqrt(2))`
= `1/sqrt(2)`
= cos 45°
APPEARS IN
संबंधित प्रश्न
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 tan `(7pi)/12`
Prove that cos(π + θ) = − cos θ
Prove that sin(30° + θ) + cos(60° + θ) = cos θ
Prove that sin(n + 1) θ sin(n – 1) θ + cos(n + 1) θ cos(n – 1)θ = cos 2θ, n ∈ Z
Prove that cos(A + B) cos(A – B) = cos2A – sin2B = cos2B – sin2A
If cos(α – β) + cos(β – γ) + cos(γ – α) = `- 3/2`, then prove that cos α + cos β + cos γ = sin α + sin β + sin γ = 0
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
Find the value of cos 2A, A lies in the first quadrant, when cos A = `15/17`
If θ is an acute angle, then find `cos (pi/4 + theta/2)`, when sin θ = `8/9`
Prove that cos 5θ = 16 cos5θ – 20 cos3θ + 5 cos θ
Prove that sin 4α = `4 tan alpha (1 - tan^2alpha)/(1 + tan^2 alpha)^2`
Express the following as a product
sin 75° sin 35°
Prove that cos(30° – A) cos(30° + A) + cos(45° – A) cos(45° + A) = `cos 2"A" + 1/4`
If A + B + C = 180°, prove that sin2A + sin2B + sin2C = 2 + 2 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:
If `pi < 2theta < (3pi)/2`, then `sqrt(2 + sqrt(2 + 2cos4theta)` equals to
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) =
