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If sinA=4/5 solve sin2A, cos2A and tan2A​

User Niveathika
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1 Answer

21 votes
21 votes

Explanation:

1) if m(∠A)∈[0;90°), then


cos(A)=√(1-sin^2A) =(3)/(5);


sin2A=2sinAcosA=2*(3)/(5) *(4)/(5)=(24)/(25);


cos2A=cos^2A-sin^2A=(9)/(25)-(16)/(25)=-(7)/(25);


tan2A=(sin2A)/(cos2A)=-((24)/(25))/((7)/(25))=-(24)/(7).

2) if m∠A∈[90°;180°), then

cos(A)=-0.6;

sin2A=-0.96;

cos2A=-0.28;

tan2A=-24/7.

User Snoone
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