2 sin2 -4 cos- sin - sin -cos 2-

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Product of Trigonometric Ratios in Terms of Their Sum

2 sin2 β+4 co...

Question

$2 s i n^{2} β + 4 c o s (α + β) s i n α s i n β + c o s 2 (α + β) =$

s i n 2 α

No worries! We‘ve got your back. Try BYJU‘S free classes today!

c o s 2 β

No worries! We‘ve got your back. Try BYJU‘S free classes today!

c o s 2 α

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

s i n 2 β

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

Similar questions

α

β

π < α - β < 2 π

sin α + sin β = - \frac{21}{65}

cos α + cos β = \frac{17}{65}

cos \frac{α - β}{2}

L e t α, β b e s u c h t h a t π < α - β < 3 π . I f s i n α + s i n β = \frac{21}{65} a n d c o s α + c o s β = - \frac{27}{65}, t h e n t h e v a l u e o f c o s \frac{α - β}{2} i s

α

β

c o s^{2} α + c o s^{2} β = 3 / 2

α

β

α + β

If $s i n α, s i n β, c o s α$ are in GP, then roots of equation $x^{2} s i n β + 2 x c o s β + s i n β = 0$ are

\frac{cos α}{cos θ} + \frac{sin α}{sin θ} = \frac{cos β}{cos θ} + \frac{sin β}{sin θ} = 1

α

β

π

\frac{cos α cos β}{{cos}^{2} θ} + \frac{sin α sin β}{{sin}^{2} θ} + 1 = 0

Join BYJU'S Learning Program