Hi,
I have heard its possible to find the sine and cosine of some common angles without calculator.
Can anyone explain how please?
eg.
cos pi/4 = 1/√2
sin pi/4 = 1/√2
cos 5pi/4 = -1/√2
cos pi/3 = 1/2
sin pi/3 = √3/2
Thanks in advance
Hi,
I have heard its possible to find the sine and cosine of some common angles without calculator.
Can anyone explain how please?
eg.
cos pi/4 = 1/√2
sin pi/4 = 1/√2
cos 5pi/4 = -1/√2
cos pi/3 = 1/2
sin pi/3 = √3/2
Thanks in advance
Hello, ScorpFire!
I have heard its possible to find the sine and cosine of some common angles without calculator.
Can anyone explain how please?
For example: .$\displaystyle \begin{array}{cccccccccccc}\sin\frac{\pi}{4} \:=\:\frac{1}{\sqrt{2}} && \sin\frac{\pi}{3} \:=\:\frac{\sqrt{3}}{2} && \sin\frac{\pi}{6} \:=\:\frac{1}{2} \\ \\[-3mm] \cos\frac{\pi}{4} \:=\:\frac{1}{\sqrt{2}} && \cos\frac{\pi}{3} \:=\:\frac{1}{2} && \cos\frac{\pi}{6} \:=\:\frac{\sqrt{3}}{2} \end{array}$
$\displaystyle \text{For }\theta = \tfrac{\pi}{4}\:(45^o)$, consider an isosceles right triangle.
Let the equal sides equal 1.Code:* * * * * h 1 * * * * * 45 * * * * * * 1
Pythagorus says the hypotenuse is $\displaystyle \sqrt{2}.$
We have: .$\displaystyle \begin{Bmatrix} opp &=& 1 \\ adj &=& 1 \\ hyp &=& \sqrt{2}\end{Bmatrix}$
And you can write the trig values of $\displaystyle \tfrac{\pi}{4}$
Memorize "one, one, square-root-of-two".
$\displaystyle \text{For }\theta =\tfrac{\pi}{3}\:(60^o)$, consider an equilateral triangle
. . with side length 2. .Draw an altitude.
We have: .$\displaystyle adj = 1,\:hyp = 2$Code:* *|* * | * 2 * |y * * | * * 60 | * * * * * * : 1 : 1 :
Pythagorus says: $\displaystyle opp = \sqrt{3}$
And you can write the trig values for $\displaystyle \tfrac{\pi}{3}$
For $\displaystyle \theta = \tfrac{\pi}{6}\;(30^o)$, turn the above right triangle on its side.
We have: .$\displaystyle \begin{Bmatrix}opp &=& 1 \\ adj &=& \sqrt{3} \\ hyp &=& 2\end{Bmatrix}$Code:* 2 * * * * 1 * 30 * * * *_ * * √3
And you can write the trig values for $\displaystyle \tfrac{\pi}{6}$
For both diagrams, memorize "one, two, square-root-of-three".
Be careful! .Remember that:
. . The shortest side, 1, is opposite the smallest angle, 30^{o}.
. . The longest side, 2, is opposite the largest angle, 90^{o}.
Take a point on the unit circle. The cosine of the corresponding angle is simply the x-coordinate (adjacent over hypotenuse). The sine of that angle is the y-coordinate. That's how I usually find sine/cosine in my head by just visualizing the unit circle.
There are formulae for finding sin(A+B) and cos(A+B) etc... using these one can extend the values for other angles. For example, sin2A = 2sinAcosA, using which one can find sin(15 degrees) = sqrt(2 - sqrt(3))/2.
Salahuddin
Maths online