Trigonometry - Trigonometric Functions

If you want to roughly approximate an EKG (pulse/heart beat diagram) of a person with a pulse of beats per minute using a sine function, what would be equal to in the following equation?

, where is measured in seconds

$3\pi$

$\pi$

$2\pi$

$4\pi$

$5\pi$

Explanation

Firstly, we need to realize that because time in this function is measured in seconds and we need to produce a function that approximates heartbeats in seconds (or periods in seconds), our frequency is...

$\frac{90}{60} = \frac{3}{2}$ beats per second.

We can take the reciprocal from here to get our period for a single heartbeat:

$(\frac{3}{2})^{-1} = \frac{2}{3}$ seconds.

Finally, since we know that $\frac{2\pi}{b}$ must be our period, we can solve for using algebra.

$\frac{2\pi}{b} = \frac{2}{3}$

$2b = 6\pi$

$b = \frac{6\pi}{2} = 3\pi$

If sin 26o = t, find the value of cos 52o in terms of t.

Explanation

Since $52^{\circ}=2\times 26^{\circ}$ , we can use a double angle formula:

$cos\ 2x=1-2sin^2x\Rightarrow cos\ 52^{\circ}=1-2sin^2\ (26)^{\circ}$

Substituting $sin \26^{\circ}=t$ , we get $cos\ 52^{\circ}=1-2sin^2\ (26)^{\circ}=1-2t^2$ .

If $sin^2x+2cos^2x+3(sin\ x)(cos\ x)=3$ , give the value of $cot\ x$ .

$cot\ x=1\ or\ cot\ x=2$

$cot\ x=1$

$cot\ x=-1\ or\ cot\ x=2$

$cot\ x=2$

$cot\ x=1\ or\ cot\ x=-2$

Explanation

$sin^2x+2cos^2x+3(sin\ x)(cos\ x)=3$

$\Rightarrow sin^2x+(cos^2x+cos^2x)+3(sin\ x)(cos\ x)=3$

Use the identity .

$\Rightarrow 1+cos^2x+3(sin\ x)(cos\ x)=3\Rightarrow cos^2x+3(sin\ x)(cos\ x)=3-1=2$

Now we should divide both sides by :

$\Rightarrow \frac{cos^2x}{sin^2x}+\frac{3(sin\ x)(cos\ x)}{sin^2x}=\frac{2}{sin^2x}$

We can use the identity $\frac{1}{sin^2x}=1+cot^2x$ .

$\Rightarrow cot^2x+3cot\ x=2(1+cot^2x)$

$\Rightarrow cot^2x+3cot\ x=2+2cot^2x$

$\Rightarrow 2cot^2x-cot^2x-3cot\ x+2=0$

$\Rightarrow cot^2x-3cot\ x+2=0$

$\Rightarrow (cot\ x-1)(cot\ x-2)=0$

$cot\ x-1=0\Rightarrow cot\ x=1$

$cot\ x-2=0\Rightarrow cot\ x=2$

Rt_triangle_letters

In this figure, if angle $c=90^\circ$ , side , and side , what is the measure of angle ?

$41.8^\circ$

$48.2^\circ$

Undefined

$90^\circ$

$\frac{2}{3}^\circ$

Explanation

Since $c=90^\circ$ , we know we are working with a right triangle.

That means that $\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}$ .

In this problem, that would be:

$\sin(a)=\frac{\text{Z}}{\text{X}}$

Plug in our given values:

$\sin(a)=\frac{8}{12}$

$\sin(a)=\frac{2}{3}$

$a=\sin^{-1}(\frac{2}{3})$

$a=41.8^\circ$

If $sin\ x+\frac{1}{sin\ x}=2$ , find the value of .

Explanation

$sin\ x+\frac{1}{sin\ x}=2\Rightarrow \frac{sin^2x+1}{sin\ x}=2$

$\Rightarrow sin^2x+1=2sin\ x$

$\Rightarrow sin^2x-2sin\ x+1$

$\Rightarrow (sinx-1)^2=0$

$\Rightarrow sin\ x-1=0$

$\Rightarrow sin\ x=1$

Therefore, .

If $x=\frac{\pi}{6}$ , find the value of $\frac{1+cos\ x}{sin^3x}$ .

$8+4\sqrt{3}$

$8+4\sqrt{2}$

$4+4\sqrt{3}$

$-8+4\sqrt{3}$

$8-4\sqrt{3}$

Explanation

$x=\frac{\pi}{6}\Rightarrow \frac{{1+cos\ x}}{}{sin^3x}=\frac{{1+cos\ (\pi/6)}}{sin^3(\pi/6)}=\frac{1+\frac{\sqrt{3}}{2}}{(\frac{1}{2})^3}=4(2+\sqrt{3})=8+4\sqrt{3}$

Find the value of the following expression:

$sin^2 (\frac{\pi}{15})+\frac{1}{sec^2(\frac{\pi}{15})}$

Explanation

We know that $sec\ x= \frac{1}{cos\ x}$ .

$sin^2 (\frac{\pi}{15})+\frac{1}{sec^2(\frac{\pi}{15})}=sin^2 (\frac{\pi}{15})+\frac{1}{\frac{1}{cos^2(\frac{\pi}{15})}}=sin^2 (\frac{\pi}{15})+cos^2 (\frac{\pi}{15})=1$