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Question : 22 of 160
Marks:
+1,
-0
Solution:
‌sin‌310∘+sin‌350∘−sin‌370∘‌⇒‌‌sin‌3x=3sin‌x−4sin‌3x‌⇒‌‌sin‌3x=‌(3sin‌x−sin‌3x)‌⇒‌‌‌[3sin‌10∘−sin‌30∘+3sin‌50∘−sin‌150∘−3sin‌70∘+sin‌210∘]‌⇒‌‌‌[3sin‌10∘+3sin‌50∘−3sin‌70∘−‌−‌−‌]‌⇒‌[sin‌10∘+sin‌50∘−sin‌70∘−‌]‌⇒‌[2sin‌30∘‌cos‌20∘−sin‌70∘−‌]‌⇒‌[cos‌20∘−sin‌70∘−‌]‌⇒‌[sin‌70∘−sin‌70∘−‌]=−‌
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