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Question : 23 of 160
Marks:
+1,
-0
Solution:
Given,
‌| √2‌cos‌45∘+cos‌56∘+cos‌58∘−cos‌66∘ |
| √2‌cos‌28∘‌cos‌29∘‌s‌i‌n‌33∘ |
‌=‌| √2()+cos‌56∘+cos‌58∘−cos‌66∘ |
| √2‌cos‌28∘‌cos‌29∘‌s‌i‌n‌33∘ |
=‌‌| (1−cos‌66∘)+2‌cos‌57∘‌cos‌1∘ |
| √2‌cos‌28∘‌cos‌29∘‌s‌i‌n‌33∘ |
=‌‌| 2sin‌233∘+2sin‌33∘‌cos‌1∘ |
| √2‌cos‌28∘‌cos‌29∘‌s‌i‌n‌33∘ |
=‌2sin‌33∘(‌| sin‌33∘+cos‌1∘ |
| √2‌cos‌28∘‌cos‌29∘‌s‌i‌n‌33∘ |
)=‌√2(‌| cos‌57∘+cos‌1∘ |
| cos‌28∘‌cos‌29∘ |
)=√2(‌| 2‌cos‌29∘‌cos‌28∘ |
| cos‌28∘‌cos‌29∘ |
)=‌2√2
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