# Math Help - Maths- Trigonometry Exact Values

1. ## Maths- Trigonometry Exact Values

Hello, I just started this topic and is already struggling with a few questions, that I need help kindly need help with, as I am unable to get the correct answer.
Question 1) Find the exact value of the following:
a. Cos315
I thought the method to this was to subtract it with the nearest degree.
Therefore being 270=315-270
=cos45.
And according to the table I was given cos 45 is √2/2.
http://moodle.tbaisd.org/file.php/87...atio_table.gif
Similarly
b. Sin240
I thought the answer would be 270(nearest)-240=30
Thus Sin30
=-√1/2

2. ## Re: Maths- Trigonometry Exact Values

for cos(315)=cos(360-45)=cos(45)=1/sqrt(2) as cos(360-theta)=cos(360)cos(theta)+sin(360)sin(theta)
now cos(360)=1 and sin(360)=0 and theta for above is=45(in degree)
for sin(240)=sin(360-120)=sin(360)cos(120)-cos(360)sin(120)
now using cos(360)=1 and sin(360)=0
we get -sin(120)=-sin(180-60)=-sin(60)=-(sqrt(3)/2)
as sin(180-60)=sin(180)cos(60)-cos(180)sin(60)
use sin(180)=0 and cos(180)=-1
so you get sin(180-60)=sin(60)

3. ## Re: Maths- Trigonometry Exact Values

Originally Posted by Qutyberry
Hello, I just started this topic and is already struggling with a few questions, that I need help kindly need help with, as I am unable to get the correct answer.
Question 1) Find the exact value of the following:
a. Cos315
I thought the method to this was to subtract it with the nearest degree.
Therefore being 270=315-270
=cos45.
And according to the table I was given cos 45 is √2/2.
http://moodle.tbaisd.org/file.php/87...atio_table.gif
Generally speaking, one should not attempt trigonometry until one has mastered arithmetic- including square roots. By definition of "square root", $(\sqrt{a})(\sqrt{a})= a$ so that $2= (\sqrt{2})(\sqrt{2})$. Then $\frac{\sqrt{2}}{2}= \frac{\sqrt{2}}{(\sqrt{2})(\sqrt{2})}= \frac{1}{\sqrt{2}}$.

Have you never learned to "rationalize the denominator"?
By the way, you don't have to memorize these or look then up. 45 degrees is exactly half of 90 degrees so if a right triangle has one angle 45 degrees the other is also 45 degrees- we have an "isosceles triangle"- the two legs are of the same length. If we call that length "x", by the Pythagorean theorem, the hypotenuse has length $\sqrt{x^2+ x^2}= \sqrt{2x^2}= x\sqrt{2}$ (x is positive). The sin(45) is "opposite side over hypotenuse"= $\frac{x}{x\sqrt{2}}= \frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2}$.

Similarly
b. Sin240
I thought the answer would be 270(nearest)-240=30
Thus Sin30
=-√1/2
I don't know where you got " $sin(30)= -\sqrt{1/2}$". That's clearly wrong. In the first place, angles between 0 and 90 degrees correspond to actual right triangles which have only positive sides. And you already know that $\sqrt{1/2}$ is the sine of 45 degrees, not 30 degrees. All the ratios, and so all the trig functions, are positive in the first quadrant. The easiest way to get "sin(30)" (or cos(30) or sin(60) or cos(60)) is to look at an equilateral triangle. An equilateral triangle has all sides of equal length and all angles of equal measure, so 180/3= 60. Now draw a perpendicular from one angle to the opposite side. It is easy to prove (using "congruent triangles") that it also bisects both the angle and the opposite side. So if we call one side of the equilateral triangle "x", then we have two right triangles with hypotenuse of length x and one leg of length x/2. The other side has length $\sqrt{x^2- (x/2)^2}= \sqrt{x^2- x^2/4}= \sqrt{3x^2/4}= x\sqrt{3}/2$. That is, the side opposite the 30 degree angle has length $x\sqrt{3}/2$ so $sin(30)= \frac{x\sqrt{3}/2}{x}= \frac{\sqrt{3}}{2}$. Now, 240 is in the third quadrant (it is between 180 and 270 degrees) so the "opposite side" is downward (negative) while the hypotenuse is always positive: $sin(240)= -\sqrt{3}/2$.