(sin t + cos t)^2?

Question

Can someone give me the answer to it and the process how you got the answer? Alos, what is the problem called, the process. Right now I am learning trig and need future reference.

Good answers guys...Thanks....

Answer 1

You can use the basic algebra fact that
(x + y)^2 = x^2 + 2 x y + y^2

In this case, with x = sin t and y = cos t, you find
(sin t + cos t)^2 = (sin t)^2 + 2 sin t cos t + (cos t)^2

Note: mathematicians don't like parentheses so they write sin^2 t instead of (sin t)^2, etc.

It is a basic fact in trigonometry that the square of the sine plus the square of the cosine of the same angle are always equal to 1. In symbols: sin^2 t + cos^2 t = 1.

Therefore,
(sin t + cos t)^2 = (sin t)^2 + (cos t)^2 + 2 sin t cos t
... = 1 + 2 sin t cos t

Finally, the product (2 sin t cos t) is special because it is precisely the sine of *twice* the angle, that is,

2 sin t cos t = sin (2t)

Therefore, the expression simplifies as

(sin t + cos t)^2 = 1 + sin (2t)

as many others already pointed out...

Answer 2

The process is squaring a binomial. It follows the form:

(a+b)^2 = a^2 + 2ab + b^2

therefore,

(sin t + cos t)^2 = sin^2 t + 2 sin t cos t + cos ^2 t

however, by definition

sin ^2 t + cos ^2 t = 1

so, the result is

(sin t + cos t)^2 = 1 + 2 sin t cos t

Answer 3

=sin^2 t + 2 sintcost + cos^2 t
=1+2sintcost
=1+sin(2t)

Answer 4

(sin t + cos t)^2
=(sin t + cos t)(sin t + cos t)
=(sin t)^2 + 2.sin t.cos t + (cos t)^2
=1 + 2.sin t.cos t
=1 + sin 2t
[Note: (sin t)^2 + (cos t)^2 = 1 and 2.sin t.cos t = sin 2t]

Answer 5

=sin^2(t) + cos^2(t) + 2sin(t)cos(t)
=1+ 2sin(t)cos(t) as sin^2(t) + cos^2(t)=1

Answer 6

the first answer is right, and they are called trigonometric identities for your future reference.

Answer 7

=sin^2(t) + cos^2(t) + 2sin(t)cos(t)
=1+ 2sin(t)cos(t) bcoz sin^2(t) + cos^2(t)=1

Answer 8

sin^2t+cos^2t+2sintcost
=>1+2sintcost
=>1+sin2t

Answer 9

What is the question ???????