Trigonometry

[Guest]

Show that the equation 3sin^(2)θ-2cos^(2)θ=1
can be written as 5sin^(2)θ=3

[Math Tutor]

sin²t + cos²t = 1
so
cos²t = 1 - sin²t

Can you write the rest?

[Guest]

sinx+cosx=1

[leesajohnson]

sin²t + cos²t = 1

[Guest]

You need to do the sum........not just write out the trig. identity.....

Show that the equation 3sin^2t-2cos^2t=1
can be written as 5sin^2t=3

Cos^2t+Sin^2t=1 ....>>...Cos^2t=1-Sin^2t

So ... 2cos^2t = 2(1-Sin^2t)

and ... 3sin^2t-2cos^2t = 1 ...same as... 3sin^2t-2+2Sin^2t = 1 ...same as... 5Sin^2t = 3

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