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Sin^2A-cos^2B=sin2^B-cos^2A​

Sin^2A-cos^2B=sin2^B-cos^2A​
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3 votes
Sin^2A-cos^2B=sin2^B-cos^2A​

User Not Link
by
8.5k points

1 Answer

1 vote

Answer:

see explanation

Explanation:

Using the trig identity

sin²x + cos²x = 1 , then

sin²x = 1 - cos²x

cos²x = 1 - sin²x

Consider the left side

sin²A - cos²B

= (1 - cos²A) - (1 - sin²B) ← distribute parenthesis

= 1 - cos²A - 1 + sin²B

= sin²B - cos²A

= right side , thus verified

User Agyei
by
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