Advertisements
Advertisements
Question
Prove that sin(45° + θ) – sin(45° – θ) = `sqrt(2) sin θ`
Advertisements
Solution
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
RELATED QUESTIONS
Find the values of tan(1050°)
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:
sec θ = `13/5`, θ lies in the IV quadrant
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)
Prove that cos 8θ cos 2θ = cos25θ – sin23θ
Show that tan(45° − A) = `(1 - tan "A")/(1 + tan "A")`
Show that `cot(7 1^circ/2) = sqrt(2) + sqrt(3) + sqrt(4) + sqrt(6)`
Prove that `32(sqrt(3)) sin pi/48 cos pi/48 cos pi/24 cos pi/12 cos pi/6` = 3
Express the following as a product
sin 50° + sin 40°
Express the following as a product
cos 35° – cos 75°
Show that `((cos theta -cos 3theta)(sin 8theta + sin 2theta))/((sin 5theta - sin theta) (cos 4theta - cos 6theta))` = 1
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 sin2A + sin2B − sin2C = 2 sin A sin B cos C
If A + B + C = 180°, prove that sin A + sin B + sin C = `4 cos "A"/2 cos "B"/2 cos "C"/2`
If ∆ABC is a right triangle and if ∠A = `pi/2` then prove that cos2 B + cos2 C = 1
If ∆ABC is a right triangle and if ∠A = `pi/2` then prove that sin2 B + sin2 C = 1
If ∆ABC is a right triangle and if ∠A = `pi/2` then prove that cos B – cos C = `- 1 + 2sqrt(2) cos "B"/2 sin "C"/2`
Choose the correct alternative:
If cos 28° + sin 28° = k3, then cos 17° is equal to
Choose the correct alternative:
cos 1° + cos 2° + cos 3° + ... + cos 179° =
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) =
