Question

find the value of Sin 215degree+Sin 225degree+sin 265degree+sin275degree is
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Answer 1

To Find :-

• sin215° + sin225° + sin265° + sin275°

Formula used :-

• sin(180+θ) = - sinθ (3rd Quadrant).
• sin(360 - θ) = - sinθ (4th Quadrant).
• sinC + sinD = 2*sin(C+D/2) * cos(C-D/2)

Solution :-

→ sin215° = sin(180° + 35°)

→ sin(180+θ) = - sinθ

So,

→ sin215° = - sin35°

Similarly,

→ sin225° = sin(180° + 45°)

→ sin(180+θ) = - sinθ

So,

→ sin225° = - sin45°

And,

→ sin265° = sin(360° - 95°)

→ sin(360 - θ) = - sinθ

So,

→ sin265° = - sin95°

Similarly,

→ sin275° = sin(360° - 85°)

→ sin(360 - θ) = - sinθ

So,

→ sin275° = - sin85°

___________________

So,

→ sin215° + sin225° + sin265° + sin275°

Putting All Values we get,

→ - sin35° - sin45° - sin95° - sin85°

→ - [ sin35° + sin45° + sin95° + sin85°]

→ - [(sin95° + sin35°) + (sin85° + sin45°)]

using sinC + sinD = 2*sin(C+D/2) * cos(C-D/2) Now,

→ - [ {2*sin(95+35)/2 * cos(95 - 35)/2 } + {2*sin(85+45)/2 * cos(85 - 45)/2 }]

→ - [ (2*sin65°*cos30) + (2*sin65°*cos20°) ]

→ - [ {2 * sin65° * (1/2)} + (2*sin65°*cos20°) ]

→ - [ sin65° + 2*sin65°*cos20° ]

→ (-sin65°) [ 1 + 2cos20°] (Ans).

Answer 2

QUESTION :

Find the value of Sin 215° + Sin 225° + sin 265° + sin275° .

SOLUTION :

We know that :

Sin is + ve in the 1 st and 2 nd quadrant

Sin is - ve in the 3rd and 4th quadrant.

Sin 215 ° = Sin ( 180 +35 ) ° i.e - sin ( 35°)

Sin 225 ° = Sin ( 180 +45 ) ° i.e - sin ( 45°)

Similarly sin 265° = - sin 95

Sin 275 ° = - sin 85

Now using Sin C + Sin D Fromula :

=> - [ Sin 85 + Sin 95 + sin 35 + sin 45 ]

=> - sin 65 { 1 + cos2• } ..........(A)