Answer to Question #167794 in Civil and Environmental Engineering for eyel

Question

A.     Solve the following problems. Write your answer on a sheet of paper.

1.      Given that sin α =  and sin β = , find 𝑡𝑎𝑛 (α + β) if both α and β are in QIV.

2.      Given that csc A = , A in QI, and sec B = , sin B < 0, find

a.      cos (A – B)

b.     tan (A – B)

3.      If tan (x + y) =  and tan y = , what is tan x?

4.      The point (9, –5) lies on the terminal side of the angle 𝜃 in standard position. Find (sin 𝜃 + cos 𝜃).

B.      Solve the following problems. Write your answer on a sheet of paper.

1.      If cos t = , what is cos 2t?

2.      Use half – angle identities to find the exact value of tan 22.5⁰ and sin 15⁰.

Answer 1

1.

$\sin \alpha=a, -1\leq a\leq0$

$\sin\beta=b, -1\leq a\leq0$

$\cos \alpha=\sqrt{1-\sin^2 \alpha}=\sqrt{1-a^2}$

$\cos\beta =\sqrt{1-\sin^2 \beta}=\sqrt{1-b^2}$

$\tan(\alpha+\beta)=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$

$=\dfrac{\dfrac{a}{\sqrt{1-a^2}}+\dfrac{b}{\sqrt{1-b^2}}}{1-\dfrac{a}{\sqrt{1-a^2}}\dfrac{b}{\sqrt{1-b^2}}}$

$=\dfrac{a\sqrt{1-b^2}+b\sqrt{1-a^2}}{\sqrt{1-a^2}\sqrt{1-b^2}-ab}$

$\tan(\alpha+\beta)=\dfrac{a\sqrt{1-b^2}+b\sqrt{1-a^2}}{\sqrt{1-a^2}\sqrt{1-b^2}-ab}$

2.

$\csc A=\dfrac{1}{\sin A}=a, a\geq 1$

$\sin A=\dfrac{1}{a},$

$\cos A=\sqrt{1-\sin^2 A}=\sqrt{1-(\dfrac{1}{a})^2}=\dfrac{\sqrt{a^2-1}}{a}$

$\sec B=\dfrac{1}{\cos B}=b, |b|\geq 1$

$\cos B=\dfrac{1}{b}$

$\sin B=-\sqrt{1-\cos^2 B}=-\sqrt{1-(\dfrac{1}{b})^2}=-\dfrac{\sqrt{b^2-1}}{|b|}$

a.

$\cos(A-B)=\cos A\cos B+\sin A\sin B$

$=\dfrac{\sqrt{a^2-1}}{a}(\dfrac{1}{b})+\dfrac{1}{a}(-\dfrac{\sqrt{b^2-1}}{|b|})$

$b\geq1, a\geq1:\cos(A-B)=\dfrac{\sqrt{a^2-1}-\sqrt{b^2-1}}{ab}$

$b\leq-1, a\geq1:\cos(A-B)=\dfrac{\sqrt{a^2-1}+\sqrt{b^2-1}}{ab}$

b.

$\tan(A-B)=\dfrac{\sin(A-B)}{\cos(a-B)}$

$\sin(A-B)=\sin A\cos B-\cos A\sin B$

$=\dfrac{1}{a}(\dfrac{1}{b})-\dfrac{\sqrt{a^2-1}}{a}(-\dfrac{\sqrt{b^2-1}}{|b|})$

$b\geq1, a\geq1:\cos(A-B)=\dfrac{\sqrt{a^2-1}-\sqrt{b^2-1}}{ab}$

$\sin(A-B)=\dfrac{1+\sqrt{(a^2-1)(b^2-1)}}{ab}$

$\tan(A-B)=\dfrac{1+\sqrt{(a^2-1)(b^2-1)}}{\sqrt{a^2-1}-\sqrt{b^2-1}}$

$b\leq-1, a\geq1:\cos(A-B)=\dfrac{\sqrt{a^2-1}+\sqrt{b^2-1}}{ab}$

$\sin(A-B)=\dfrac{1-\sqrt{(a^2-1)(b^2-1)}}{ab}$

$\tan(A-B)=\dfrac{1-\sqrt{(a^2-1)(b^2-1)}}{\sqrt{a^2-1}+\sqrt{b^2-1}}$

3.

$\tan(x+y)=\dfrac{\tan x+\tan y}{1-\tan x \tan y}$

$\tan (x+y)-\tan x \tan (x+y)\tan y=\tan x+\tan y$

$\tan x=\dfrac{\tan (x+y)-\tan y}{1+\tan (x+y) \tan y}$

Given $\tan(x+y)=a, \tan y=b$

$\tan x=\dfrac{a-b}{1+ab}$

4.

$\sqrt{(9)^2+(-5)^2}=\sqrt{106}$

$\sin \theta=\dfrac{-5}{\sqrt{106}}, \cos\theta=\dfrac{9}{\sqrt{106}}$

$\sin \theta+\cos \theta =-\dfrac{5}{\sqrt{106}}+\dfrac{9}{\sqrt{106}}=\dfrac{4}{\sqrt{106}}$

$\sin \theta+\cos \theta =\dfrac{4}{\sqrt{106}}$

B.

1.     

$\cos2t=2\cos^2t-1$

Given $\cos t=a$

$\cos2t=2a^2-1$

2.

$\sin 15\degree=\sqrt{\dfrac{1-\cos 30\degree}{2}}=\sqrt{\dfrac{1-\dfrac{\sqrt{3}}{2}}{2}}$

$\sin 15\degree=\dfrac{\sqrt{2-\sqrt{3}}}{2}$

$\tan 22.5\degree=\sqrt{\dfrac{1-\cos 45\degree}{1+\cos45\degree}}=\sqrt{\dfrac{1-\dfrac{\sqrt{2}}{2}}{1+\dfrac{\sqrt{2}}{2}}}$

$=\sqrt{\dfrac{2-\sqrt{2}}{2+\sqrt{2}}}=\dfrac{2-\sqrt{2}}{\sqrt{2}}=\sqrt{2}-1$

$\tan 22.5\degree=\sqrt{2}-1$