Question

I need answer quickly in step by step answer ​

Answer 1

L.H.S.

$2 \sin( \frac{10 + 20}{2} ) \cos( \frac{20 - 10}{2} ) + 2 \sin( \frac{40 + 50}{2} ) \cos( \frac{50 - 40}{2} )$

= 2sin15cos5+2sin45cos5
[ using sin C+sin D= 2sin C+D/2 cos C-D/2 for sin10+sin20 & sin40+sin50]

= 2cos5 (sin15+sin45)

= 2cos5 (2sin30cos15) [ using sin C+sin D= 2sin C+D/2 cos C-D/2 ]

= 2cos5 (2 x 1/2 x cos15)

= 2cos5 cos15

R.H.S. = sin70+sin80

= 2sin75cos5 [ using sin C+sin D= 2sin C+D/2 cos C-D/2 ]

sin75 = sin(90-15) = cos 15

L.H.S = 2cos5 cos15

R.H.S. = 2cos15 cos5 [ since, sin75 = cos15 ]

hope this helps you

Answer 2

Answer:

step-by-step explanation:

Required to prove:

sin 10° + sin 20° + sin 40° + sin 50° = sin 70° + sin 80°

As we know that,

sin C + sin D = 2 Sin $\frac{(C+D)}{2}$• Cos $\frac{(C-D)}{2}$

Applying this identity,

L.H.S

= (Sin 10° + sin 20 °) + (sin 40° + sin 50° )

= 2 sin $\frac{(10°+20°)}{2}$•cos$\frac{(20°-10°)}{2}$ + 2 sin $\frac{(40°+50°)}{2}$•cos $\frac{(50°-40°)}{2}$

= 2 sin 15°• cos 5° + 2 sin 90°• cos 5°

= 2 sin 15°•cos 5° + 2 sin 45° cos 5°

Taking 2cos 5° as common, we get

= 2 Cos 5° ( sin 15° + sin 45° )

= 2 cos 5° ( 2 sin$\frac{(45°+15°)}{2}$•cos $\frac{(45°-15°)}{2}$ )

= 2 cos 5° ( 2 sin 30°•cos 15° )

= 2 cos 5° ( 2× 1/2 • cos 15° )

= 2 cos 5°• cos 15°

= 2 cos 5°• cos ( 90°-75° )

now,

we know that,

cos ( 90° - @ ) = sin @

so, further L.H.s

= 2 cos 5°•sin 75°

= 2 sin $\frac{(80°+70°)}{2}$• cos $\frac{(80°-70°)}{2}$

= sin 70° + 80°

= L.H.S

Hence,

proved..

ANOTHER APPROACH :-

L.H.S

= (Sin 10° + sin 20 °) + (sin 40° + sin 50° )

= 2 sin $\frac{(10°+20°)}{2}$•cos$\frac{(20°-10°)}{2}$ + 2 sin $\frac{(40°+50°)}{2}$•cos $\frac{(50°-40°)}{2}$

= 2 sin 15°• cos 5° + 2 sin 90°• cos 5°

= 2 sin 15°•cos 5° + 2 sin 45° cos 5°

Taking 2cos 5° as common, we get

= 2 Cos 5° ( sin 15° + sin 45° )

= 2 cos 5° ( 2 sin$\frac{(45°+15°)}{2}$•cos $\frac{(45°-15°)}{2}$ )

= 2 cos 5° ( 2 sin 30°•cos 15° )

= 2 cos 5° ( 2× 1/2 • cos 15° )

= 2 cos 5°• cos 15°

Now,

R.H.S

= sin 70° + sin 80°

= 2 sin $\frac{(80°+70°)}{2}$• cos $\frac{(80°-70°)}{2}$
= 2 sin 75°• cos 5°

= 2 sin (90°- 15°)• cos5°

but,

sin (90°- @ ) = cos @

so, further

R.H.S

= 2 cos 15°• cos 5°

Thus,

L.H.S = R.H.S

Hence, proved