Question

If sinA=3/5 and cosB=5/13 and if A and B are measures of two angles in Quadrant I, find the exact value of the following functions.

cotB =

sin2A=

3)cos(5pi/6 + B) =

tan(A - pi/4) =

Answer 1

1.
`\cot B=(5)/(12)`

2.
`\sin2A=(24)/(25)`

3.
`\cos((5\pi)/(6)+B )=-(5)/(26)(√(3)+1)`

4.
`\tan(A-(\pi)/(4) )=-(1)/(7)}`

Explanation:

If
`\sin A=(3)/(5)` and
`\cos B=(5)/(13)`, then we can use the Pythagorean identity to find
`\cos A` and
`\sin B`.

`\sin^2A+\cos^2A=1`

`((3)/(5) )^2+\cos^2 A=1`

`(9)/(25)+\cos ^2A=1`

`\cos^2 A=1-(9)/(25)`

`\cos^2 A=(16)/(25)`

`\cos A=\pm \sqrt{(16)/(25)}`

Since A is in quadrant I,

`\cos A=\sqrt{(16)/(25)}`

`\cos A=(4)/(5)`

Also;

`\sin^2B+\cos^2B=1`

`((5)/(13) )^2+\sin^2 B=1`

`(25)/(169)+\sin ^2B=1`

`\sin^2 B=1-(25)/(169)`

`\sin^2 B=(144)/(169)`

`\sin B=\pm \sqrt{(144)/(169)}`

Since A is in quadrant I,

`\sin B=\sqrt{(144)/(169)}`

`\sin B=(12)/(13)`

This implies that;

`\cot B=(\cos B)/(\sin B)`

`\cot B=((5)/(13) )/((12)/(13) )`

`\cot B=(5)/(12)`

`\sin2A=2\sin A \cos A`

`\sin2A=2* (3)/(5) * (4)/(5)`

`\sin2A=(24)/(25)`

`\cos((5\pi)/(6)+B )=\cos((5\pi)/(6))\cos(B )-\sin((5\pi)/(6))\sin(B)`

This implies that;

`\cos((5\pi)/(6)+B )=\cos((5\pi)/(6))* (5)/(13)-\sin((5\pi)/(6))* (5)/(13)`

`\cos((5\pi)/(6)+B )=-(√(3))/(2)* (5)/(13)-(1)/(2))* (5)/(13)`

`\cos((5\pi)/(6)+B )=-(5)/(26)(√(3)+1)`

`\tan(A-(\pi)/(4) )=(\tan A-\tan (\pi)/(4) )/(1+\tan A \tan (\pi)/(4))`

But;
`\tan A=(\sin A)/(\cos A)`

`\tan A=((3)/(5) )/((4)/(5) )=(3)/(4)`

`\tan(A-(\pi)/(4) )=((3)/(4)-\tan (\pi)/(4) )/(1+(3)/(4) \tan (\pi)/(4))`

Simplify;

`\tan(A-(\pi)/(4) )=((3)/(4)-1 )/(1+(3)/(4))`

`\tan(A-(\pi)/(4) )=-(1)/(7)}`