Advertisements
Advertisements
Question
For ΔABC , prove that :
`sin((A + B)/2) = cos"C/2`
Advertisements
Solution
`sin((A + B)/2) = cos"C/2`
We know that for a triangle ΔABC
`<A + <B + <C = 180^circ`
`(<B + <A)/2 = 90^circ - (<C)/2`
`sin((A+B)/2) = sin(90^circ - C/2)`
= `cos(C/2)`
APPEARS IN
RELATED QUESTIONS
Prove the following trigonometric identities
`(1 + tan^2 theta)/(1 + cot^2 theta) = ((1 - tan theta)/(1 - cot theta))^2 = tan^2 theta`
Prove the following trigonometric identities
cosec6θ = cot6θ + 3 cot2θ cosec2θ + 1
Prove the following identities:
`1/(tan A + cot A) = cos A sin A`
Prove the following identities:
sec4 A (1 – sin4 A) – 2 tan2 A = 1
`sin^2 theta + 1/((1+tan^2 theta))=1`
If x = a sec θ + b tan θ and y = a tan θ + b sec θ, prove that (x2 – y2) = (a2 – b2).
If 5x = sec ` theta and 5/x = tan theta , " find the value of 5 "( x^2 - 1/( x^2))`
If sin θ = `11/61`, find the values of cos θ using trigonometric identity.
Prove the following identities.
`sqrt((1 + sin theta)/(1 - sin theta)) + sqrt((1 - sin theta)/(1 + sin theta))` = 2 sec θ
Prove that sin2A . tan A + cos2A . cot A + 2 sin A . cos A = tan A + cot A.
