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Some Applications Of Trigonometry

Question

Show that:
cos 38° cos 52° - sin 38° sin 52° = 0.

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Solution

LHS
= cos 38° . cos 52° - sin 38° . sin 52°
= cos (90°-52°). cos 52°-sin (90° - 52°). sin 52°
= sin 52° . cos 52° - cos 52° . sin 52°
= 0 = RHS Proved.

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Some Applications Of Trigonometry

Show that:
cos 38° cos 52° - sin 38° sin 52° = 0.

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For Daily Free Study Material Join wiredfaculty

Some Applications Of Trigonometry

Show that:cos 38° cos 52° - sin 38° sin 52° = 0.

Some More Questions From Some Applications of Trigonometry Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Show that:
cos 38° cos 52° - sin 38° sin 52° = 0.