Advertisements
Advertisements
प्रश्न
Express the following as a product
cos 65° + cos 15°
Advertisements
उत्तर
We know sin C – sin D = `2 cos ("C" + "D")/2 * cos ("C" - "D")/2`
Take C = 65°, D = 15°
cos 65° + cos 15° = `2cos((65^circ + 15^circ)/2) * cos((65^circ - 15^circ)/2)`
cos 65° + cos 15° = `2cos(80^circ/2) * cos(50^circ/2)`
cos 65° + cos 15° = 2 cos 40° . cos 25°
APPEARS IN
संबंधित प्रश्न
Find the value of the trigonometric functions for the following:
cos θ = `- 1/2`, θ lies in the III quadrant
Prove that `(cot(180^circ + theta) sin(90^circ - theta) cos(- theta))/(sin(270^circ + theta) tan(- theta) "cosec"(360^circ + theta))` = cos2θ cotθ
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)
If sin A = `3/5` and cos B = `9/41, 0 < "A" < pi/2, 0 < "B" < pi/2`, find the value of cos(A – B)
Find sin(x – y), given that sin x = `8/17` with 0 < x < `pi/2`, and cos y = `- 24/25`, x < y < `(3pi)/2`
Prove that sin(45° + θ) – sin(45° – θ) = `sqrt(2) sin θ`
Prove that sin 105° + cos 105° = cos 45°
Prove that sin(A + B) sin(A – B) = sin2A – sin2B
Show that tan(45° − A) = `(1 - tan "A")/(1 + tan "A")`
Find the value of cos 2A, A lies in the first quadrant, when tan A `16/63`
Express the following as a sum or difference
sin 5θ sin 4θ
Express the following as a product
cos 35° – cos 75°
Prove that 1 + cos 2x + cos 4x + cos 6x = 4 cos x cos 2x cos 3x
Prove that cos(30° – A) cos(30° + A) + cos(45° – A) cos(45° + A) = `cos 2"A" + 1/4`
Prove that `(sin(4"A" - 2"B") + sin(4"B" - 2"A"))/(cos(4"A" - 2"B") + cos(4"B" - 2"A"))` = tan(A + B)
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 A + B + C = 180°, prove that sin A + sin B + sin C = `4 cos "A"/2 cos "B"/2 cos "C"/2`
If A + B + C = `pi/2`, prove the following cos 2A + cos 2B + cos 2C = 1 + 4 sin A sin B sin C
If ∆ABC is a right triangle and if ∠A = `pi/2` then prove that cos2 B + cos2 C = 1
Choose the correct alternative:
`(sin("A" - "B"))/(cos"A" cos"B") + (sin("B" - "C"))/(cos"B" cos"C") + (sin("C" - "A"))/(cos"C" cos"A")` is
