We need to use the double angle formula:

$\displaystyle sin\ 2x=2(sin\ x)(cos\ x)$

$\displaystyle sin\ x$ is known, but we need to find $\displaystyle cos \ x$:

$\displaystyle sin^2x+cos^2x=1\Rightarrow cos\ x=\pm \sqrt{1-sin^2x}$

In this problem $\displaystyle x$ is a first quadrant angle ($\displaystyle 0< x< 30^{\circ}$), so we can only use the positive value for $\displaystyle cos\ x$.

$\displaystyle cos\ x=\sqrt{1-sin^2x}=\sqrt{1-A^2}$

$\displaystyle sin\ 2x=2(sin\ x)(cos\ x)=2A\sqrt{1-A^2}$

We need to use a Pythagorean identity:

$\displaystyle sec^2x=1+tan^2x=1+(\sqrt{3})^2=1+3=4\Rightarrow sec\ x=\pm 2$

Since $\displaystyle 0^{\circ}< x< 90^{\circ}$, we can only use the positive value of $\displaystyle sec\ x$. That means $\displaystyle sec\ x=2$.

$\displaystyle cos\ x=\frac{1}{sec\ x}=\frac{1}{2}=0.5$

We need to use a Pythagorean identity:

$\displaystyle csc^2x=1+cot^2x=1+(\sqrt{3})^2=1+3=4\Rightarrow csc\ x=\pm 2$

Since $\displaystyle 0^{\circ}< x< 90^{\circ}$, we can only use the positive value for $\displaystyle csc\ x$. That means $\displaystyle csc\ x=2$.

$\displaystyle sin\ x=\frac{1}{csc\ x}=\frac{1}{2}$

Based on another Pythagorean identity we have:

$\displaystyle cos^2x+sin^2x=1\Rightarrow cos^2x=1-sin^2x=1-(\frac{1}{2})^2=1-\frac{1}{4}=\frac{3}{4}$

$\displaystyle \Rightarrow cos\ x=\pm \sqrt{(\frac{3}{4})}=\pm \frac{\sqrt{3}}{2}$

Since $\displaystyle 0^{\circ}< x< 90^{\circ}$, we can again only use the positive value for $\displaystyle cos\ x$.

$\displaystyle cos\ x=\frac{\sqrt{3}}{2}$

To solve this, start by setting up a triangle that has an angle with a cotangent of 4/3. This triangle would therefore have two legs of lengths 4 and 3, with the side of length 3 being opposite the angle in question. (Cotangent is adjacent over hypotenuse.) Note that the angle itself does not have to be solved for; we just need to find its sine. To do that, we first need to find the length of the hypotenuse using the Pythagorean Theorem:

$\displaystyle 3^{2} + 4^{2} = c^{2}$

Solving this gives a hypotenuse of length 5.  Now, the sine of this angle is opposite (i.e. the side of length 3) over hypotenuse (length 5), which gives an answer of $\displaystyle \frac{3}{5}$.

Breaking down the equation piece by piece, if we look at it as $\displaystyle f(x) = acos(bx + c) + d$:

• $\displaystyle a = d = 4$. This means both that we do have an amplitude of $\displaystyle 4$ and a minimum of $\displaystyle 4$ because $\displaystyle d - a = 4 - 4 = 0$.
• $\displaystyle c = 0$ (because it is not present), meaning that we do have no phase shift.
• $\displaystyle b = \pi$. This means that our period is $\displaystyle \frac{2\pi}{\pi} = 2$, which means that we do not have a period of 1 and that we therefore have our answer.

Remember that when a trigonometric equation is written as...

$\displaystyle f(x) = asin(bx - c) + d$

...then the phase shift is $\displaystyle c/b$ (instead of simply $\displaystyle c$.) In this problem, $\displaystyle c = \pi$ (and take careful that you do not set $\displaystyle c = -\pi$ by mistake) while $\displaystyle b = 2\pi$. Therefore, our phase shift is $\displaystyle \frac{c}{b} = \frac{\pi}{2\pi} = \frac{1}{2}$.

Examining the equation based on this form:

$\displaystyle f(x) = acos(bx + c) + d$

• $\displaystyle a = d = 3$. This means that we do have both an amplitude of $\displaystyle 3$ since $\displaystyle a = 3$ and a minimum of $\displaystyle 0$ since $\displaystyle d - a = 3 - 3 = 0$.
• $\displaystyle c = \pi$. This is not relevant until we examine $\displaystyle b$.
• $\displaystyle b = 16\pi$. Let's examine what this means step by step:
• Phase shift of $\displaystyle \frac{1}{4}$ is confirmed because our phase shift is equal to $\displaystyle \frac{c}{b} = \frac{1\pi}{16\pi} = \frac{1}{16}$.
• Frequency of $\displaystyle 8$ Hertz is confirmed. Here is why: Our period is equal to $\displaystyle \frac{2\pi}{b} = \frac{2\pi}{16\pi} = \frac{1}{8}$ seconds, which means that the frequency is $\displaystyle (\frac{1}{8})^{-1} = 8$ cycles per second.

The phase shift and period are the important points here. If our phase shift were a multiple of our period, then our $\displaystyle y$-intercept would be our maximum, which is $\displaystyle 6$, because all we are doing in that case is shifting by an amount of periods or cycles. However, $\displaystyle \frac{1}{16}$ is not a multiple of $\displaystyle \frac{1}{8}$, which means this is not the case and that we do not have a $\displaystyle y$-intercept of $\displaystyle 6$ (and it can absolutely help if you graph the function to check this.)

Firstly, we need to realize that because time in this function is measured in seconds and we need to produce a function that approximates $\displaystyle 90$ heartbeats in $\displaystyle 60$ seconds (or $\displaystyle 90$ periods in $\displaystyle 60$ seconds), our frequency is...

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

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

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

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

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

$\displaystyle 2b = 6\pi$

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

Step 1: Find the value of cos(0)

$\displaystyle cos(0)=1$, $\displaystyle x=0, y=1$

The graph of $\displaystyle cos(x)$ starts from $\displaystyle (0,1)$

First step: make the denominators equal. To do this multiply the first fraction by $\displaystyle \small \sin(105)$ and the second by $\displaystyle \small \cos(105)$. You will get $\displaystyle \small \frac{\sin^{2}(105) + \cos^{2}(105)}{\sin(105)*\cos(105)}$  . The numerator will be 1, due to the rule. To make the denominator look something decent, multiply the whole fraction by 2 . You will get $\displaystyle \small \frac{2}{2*\sin(105) *\cos(105) }$ Due to the $\displaystyle \small \sin(2x)$ rule, the denominator will                 be  $\displaystyle \small \sin(210)$, and that is $\displaystyle \small -\frac{1}{2}$. So, the last equation will be $\displaystyle \small \frac{2}{-\frac{1}{2}}$. Finally that will be equal to -4.