help needed?

Question

:) help me, or either, i'll be dead tomorrow :)

1. (sinx+cosx)^2-sin2x
2. if tan a= 1/2 a in quadran III, what is cos 2a?


helphelphelpthx:)thx:)thx:)

Answer 1

This is easy!
factor it
(sinx + cosx)(sinx + cosx) = sinx^2 + 2sincos + cos^2
sin^2+ cos ^2 =1 ( its a therom)

then its 2sincos - sin 2x
sin 2x= 2 sincos ( its a therom)
and there you go
2sincos + 2sincos = 4sincos


number two is easy just use your table of numbers

Answer 2

Problem 2.

There is an identity that says cos 2a = 2cos^2 a - 1. So, you need to find cos a.

Since tan a = 1/2, you can make a right triangle with opposite side of angle a = 1 and adjacent side of angle a = 2. Using the Pythagorean Theorem, you get the hypotenuse = sqrt (5). So, cos a = 2/sqrt (5)

Therefore, cos^2 a = 4/5. Now
cos 2a = 2cos^2 a - 1 = 2(4/5) - 1 = 8/5 - 5/5 = 3/5

Answer 3

q1)(sinx+cosx)^2=(sin x)^2+(cos x)^2+2sinx cos x
=1+2 sinx cosx [since, (sinx)^2+(cos x)^2=1]
=1+sin 2x
ans =1+sin 2x - sin 2x
=1
q2)tan a=1/2
as,cos 2a=(1- (tan a)^2)/(1+(tana)^2)

hence,cos 2a= (1-(1/2))/(1+(1/2))
=(1/2)/(3/2)
=1/3

Answer 4

problem1- ans is 1
problem2- ans is= root 3-1