# Solve the expression e = sina + cos a + sin2a + cos2a if cos a = -1/4 belongs to (pi, 3pi/2) .

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e= sina + cosa + sin2a + cos2a

**cosa = -1/4**

==> sina = sqrt(1-cos^2 a) = sqrt(1-1/16) = sqrt(15/16)

= sqrt15/4

Since in 3rd quadrant, then:

**sina == -sqrt15/4**

Let us rewrit:

we know that:

sin2a = 2sinacosa

cos2a = cos^2 a - sin^2 a

Let us substitue:

sina + cosa + 2sinacosa + cos^2 a - sin^2 a = -1/4

substitute with cosx = -1/4 and sina = -sqrt15/4

e = -sqrt15/4 - 1/4 + 2*-1/4*-sqrt15/4 + 1/16 - 15/16

= (-sqrt15-1)/4 + (2sqrt15 + 1 - 15)/16

= (-4sqrt15 - 4 + 2sqrt15 - 14)/16

= (-2sqrt15 - 18)/16

= (-sqrt15 - 9)/8

**==> e= -(sqrt15+9)/8**

First of all, before calculate sin a, we must establish to what quadrant belongs. Due to the facts from hypothesis, a is in the interval (pi, 3pi/2), we draw the conclussion that we work in the third quadrant, where the signature of sin a is minus.

cos a = -1/4

sin a = [1- (-1/4)^2]^1/2 (from the fundamental formula of trigonometry,where sin^2 a + cos^2 a = 1).

sin a = -(15)^1/2/4

In order to calculate the expression E, first we have to calculate sin 2a and cos 2a

sin 2a = sin (a+a)=sina*cosa + sina*cosa=2sina*cosa

cos 2a= cos(a+a)=cosa*cosa - sina*sina=cos^2a-sin^2a

e= sina a + cos a+ 2sina*cosa + cos^2a-sin^2a

e= -(15)^1/2/4 - 1/4 + 2*1/4*(15)^1/2/4 + 1/16 - 15/16

After finding the same denominator, which is 16, we can calculate by grouping the terms which contains (15)^1/2 together and the integer terms together.

e= [-2 -2*(15)^1/2 + (15)^1/2 - 7]/8=[-9-(15)^1/2]/8

e = sina+cosa+sin2a+cos2a +cos2a cosa= -1/4 and is in (pi,3i/2).

Both sin and cos ratios are negative for angles in 3rd quadrant.

We know sina = - sqrt(1-1/4^2) = -sqrt15/4.

sin2a = 2sinacos = 2(-1/4)(-sqr15/4) = (sqrt15)/4

cos2a = 2so^2a -1 = 2(-1/4)^2-1 = 1/8 -1 = -7/8.

Substituting in the given expression the values, we get:

Therefore e = -(sqrt15)/4 -1/4 +(sqrt15)/4 -7/8 = -1/4-7/8 = -9/4.

Therefore e = -9/4