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प्रश्न
Verify the following equalities:
cos 90° = 1 – 2sin2 45° = 2cos2 45° – 1
योग
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उत्तर
cos 90° = 0, sin 45° = `1/sqrt(2)`, cos 45° = `1/sqrt(2)`
cos 90° = 0 ...(1)
1 – 2 sin2 45° = `1 - 2(1/sqrt(2))^2`
= `1 - 2 xx 1/2`
= 1 – 1 = 0 → (2)
2 cos2 45° – 1 = `2(1/sqrt(2))^2 - 1`
= `2/2 - 1`
= `(2 - 2)/2`
= 0 → (3)
From (1), (2) and (3) we get
cos 90° = 1 – 2 sin2 45° = 2 cos2 45° – 1
Hence it is proved.
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अध्याय 6: Trigonometry - Exercise 6.2 [पृष्ठ २३२]
