Advertisements
Advertisements
Question
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`
Advertisements
Solution
A + B + C = 180°
Given A = 90°
∴ B + C = 90°
⇒ `("B" + "C")/2` = 45°
`"B"/2 + "C"/2` = 45°
R.H.S = `- 1 + 2sqrt(2) cos "B"/2 sin "C"/2`
= `1 + sqrt(2) (2cos "B"/2 sin "C"/2)`
We know that 2 cosA sinB = sin(A + B) – sin(A – B)
= `- 1 + sqrt(2) (sin (("B" + "C"))/2 - sin (("B" - "C"))/2)`
= `- 1 + sqrt(2) (sin 45^circ - sin (("B" - "C"))/2)`
= `- 1 + sqrt(2) (1/sqrt(2) - sin (("B" - "C")^2)/2)`
= `- 1 + 1 - sqrt(2) sin (("B" - "C"))/2`
= `- sqrt(2) sin (("B" - "C"))/2` .....(1)
L.H.S = cos B – cos C
= `2 sin (("B" + "C"))/2 sin (("C" - "B"))/2`
= `2 sin 45^circ sin (("C" - "B"))/2`
= `2(1/sqrt(2)) sin ((-("B" - "C"))/2)`
= `- sqrt(2) sin (("B" - "C"))/2` .....(2)
From (1) and (2)
⇒ L.H.S = R.H.S
APPEARS IN
RELATED QUESTIONS
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 cos(30° + x) = `(sqrt(3) cos x - sin x)/2`
Find a quadratic equation whose roots are sin 15° and cos 15°
Prove that sin2(A + B) – sin2(A – B) = sin2A sin2B
Show that cos2 A + cos2 B – 2 cos A cos B cos(A + B) = sin2(A + B)
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
Show that tan(45° + A) = `(1 + tan"A")/(1 - tan"A")`
Show that tan(45° − A) = `(1 - tan "A")/(1 + tan "A")`
Prove that cot(A + B) = `(cot "A" cot "B" - 1)/(cot "A" + cot "B")`
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 `sin (pi/4 - theta/2)`, when sin θ = `1/25`
Prove that cos 5θ = 16 cos5θ – 20 cos3θ + 5 cos θ
Express the following as a sum or difference
2 sin 10θ cos 2θ
Express the following as a product
sin 50° + sin 40°
Show that sin 12° sin 48° sin 54° = `1/8`
Show that `(sin 8x cos x - sin 6x cos 3x)/(cos 2x cos x - sin 3x sin 4x)` = tan 2x
Prove that sin x + sin 2x + sin 3x = sin 2x (1 + 2 cos x)
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(B + C − A) + sin(C + A − B) + sin(A + B − C) = 4 sin A sin B sin C
