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:

$cos(α+2β)=-4/5$