Answer to Question #92121 in Geometry for Lawrence Malaluan

Question

If sec α = 5/4 and cot β = 3/4, α is acute, and 1800 < β < 2700. Then, cos (α + 2β)

Answer 1

Solution

1)

$$sec(α)=1/cos(α)$$

cos(α)=4/5

from the main trigonometrical identity

$$sin^2(x)+cos^2(x)=1$$

therefore

$$sin(α)=\sqrt{1−cos^2(α)}$$

$$sin(α)=\sqrt{1−(4/5)^2}=|3/5|$$

on a condition : α is acute, therefore

sin(α)=3/5

2)

One assumes $180^{\circ}<\beta<270^{\circ}$, then cot β = 3/4.

Using the main trigonometrical identity

$$cot^2(x)+1=1/sin^2(x)$$

from this expression

$$sin(x)=\sqrt{1/(cot^2(x)+1)}$$

$$sin(β)=\sqrt{1/((3/4)^2+1)}=\sqrt{16/25}=|4/5|$$

on a condition : $180^{\circ}<\beta<270^{\circ}$, therefore

sin(β)=-4/5.

Similarly, as in the former case we will find cos(β):

$$cos(β)=\sqrt{1−sin^2(β)}$$

$$cos(β)=\sqrt{1−(-4/5)^2}=|3/5|$$

on a condition $180^{\circ}<\beta<270^{\circ}$, therefore

cos(β)=-3/5

3)

$$cos(α+2β) = cos(α)·cos(2β) - sin(α)·sin(2β)$$

$$cos(2β)=cos^2(β)-sin^2(β)=9/25-16/25=-7/25$$

$$sin(2β)=2sin(β)cos(β)=2(-4/5)(-3/5)=24/25$$

cos(2β)=-7/25, cos(α)=4/5, sin(2β)=24/25, sin(α)=3/5

$$cos(α+2β) = 4/5·(-7/25) - 3/5·24/25$$

$$cos(α+2β) = -28/125-72/125=-100/125=-4/5$$

Answer:

$cos(α+2β)=-4/5$