Advertisements
Advertisements
प्रश्न
If A, B and C are interior angles of a triangle ABC, then \[\sin \left( \frac{B + C}{2} \right) =\]
विकल्प
\[\sin \frac{A}{2}\]
\[\cos \frac{A}{2}\]
\[- \sin \frac{A}{2}\]
\[- \cos \frac{A}{2}\]
उत्तर
We know that in triangle `ABC`
`A+B+C=180°`
⇒ `B+C=180°-A`
⇒` (B+C)/2=(90°)/2-A/2`
⇒ `sin ((B+C)/2)=sin (90°-A/2)`
`"since" sin (90°-A)=cos A`
So
`sin ((B+C)/2)= cos A`
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
(cosecθ + sinθ) (cosecθ − sinθ) = cot2 θ + cos2θ
if `cot theta = 1/sqrt3` find the value of `(1 - cos^2 theta)/(2 - sin^2 theta)`
Evaluate.
sin235° + sin255°
Express the following in terms of angle between 0° and 45°:
sin 59° + tan 63°
Evaluate:
3cos80° cosec10° + 2 sin59° sec31°
Use tables to find cosine of 8° 12’
Use tables to find the acute angle θ, if the value of sin θ is 0.6525
Evaluate:
`(cos75^@)/(sin15^@) + (sin12^@)/(cos78^@) - (cos18^@)/(sin72^@)`
If 4 cos2 A – 3 = 0 and 0° ≤ A ≤ 90°, then prove that sin 3 A = 3 sin A – 4 sin3 A
If 3 cos θ = 5 sin θ, then the value of
The value of cos2 17° − sin2 73° is
If 5θ and 4θ are acute angles satisfying sin 5θ = cos 4θ, then 2 sin 3θ −\[\sqrt{3} \tan 3\theta\] is equal to
\[\frac{2 \tan 30° }{1 + \tan^2 30°}\] is equal to
Prove that :
tan5° tan25° tan30° tan65° tan85° = \[\frac{1}{\sqrt{3}}\]
Express the following in term of angles between 0° and 45° :
sin 59° + tan 63°
Find the value of the following:
`(cos 70^circ)/(sin 20^circ) + (cos 59^circ)/(sin31^circ) + cos theta/(sin(90^circ - theta))- 8cos^2 60^circ`
If cot( 90 – A ) = 1, then ∠A = ?
If sin θ + sin² θ = 1 then cos² θ + cos4 θ is equal ______.
Sin 2B = 2 sin B is true when B is equal to ______.
If A, B and C are interior angles of a ΔABC then `cos (("B + C")/2)` is equal to ______.
