User:Graemeb1967/Introduction to Calculus

The following are my solutions to Introduction to Calculus

1. Express $$\tan(\theta) \!$$ in terms of $$\sin(\theta) \!$$
 * $$\tan \theta = \frac{\sin \theta}{\cos \theta}$$

2. If $$\csc(\theta)=1/x, \!$$ then what does $$x \!$$ equal?
 * $$\csc ( \theta ) = \frac{1}{x}$$


 * $$\csc ( \theta ) = \frac{1}{sin\theta}$$


 * $$x\! = sin\theta$$

3. Prove $$\tan^2(\theta)+1=\sec^2(\theta) \!$$ using $$ \sin^2(\theta)+ \cos^2(\theta)=1 \!$$
 * $$\sin^2\theta + \cos^2\theta = 1\!$$

Dividing all terms by $$\cos^2\theta\!$$
 * $$ \frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\cos^2\theta} = \frac{1}{\cos^2\theta}$$

Now cleaning up the identities
 * $$ \frac{\sin^2\theta}{\cos^2\theta} = \tan^2\theta $$


 * $$ \frac{\cos^2\theta}{\cos^2\theta} = 1\!$$


 * $$ \frac{1}{\cos^2\theta} = \sec^2\theta$$

So finally we have
 * $$ \tan^2\theta + 1\! = \sec^2\theta\!$$

$$4. \cos(A+B)=\cos (A) \cos (B) - \sin (A)\sin (B)\!$$ '''4a. Find the double angle idenities for the cosine function using the above rule.''' Double angle formula is formed when B = A
 * $$\cos (A+A) = \cos (A) \cos (A) - \sin (A)\sin (A)\!$$


 * $$\cos (A+A) = \cos^2 (A) - \sin^2 (A)\!$$


 * $$\cos (A+A) = \cos (2A)\!$$

Therefore we have
 * $$\cos (2A) = \cos^2 (A) - \sin^2 (A)\!$$

Now to remove sin from the formula we use
 * $$\cos^2\theta + \sin^2\theta = 1\!$$

Making sin the subject
 * $$\sin^2\theta = 1 - \cos^2\theta\!$$

Substituting back into our previous equation to then find the cos identity
 * $$\cos (2A) = \cos^2 (A) - \sin^2 (A)\!$$


 * $$\cos (2A) = \cos^2 (A) - (1 - \cos^2 (A) )\!$$


 * $$\cos (2A) = \cos^2 (A) - 1 + \cos^2 (A)\!$$


 * $$\cos (2A) = 2\cos^2 (A) - 1\!$$

'''4b. Find the half angle idenities from the double angle idenities.''' We begin with this formula and rearrange it to make the 2 square the subject
 * $$\cos (2A) = 2\cos^2 (A) - 1\!$$


 * $$2\cos^2 (A) - 1 = \cos (2A) \!$$


 * $$2\cos^2 (A) = \cos (2A) + 1 \!$$

Now we half everything, notice how this now gives us half the angle.
 * $$\frac{2\cos^2 (A)}{2} = \frac{\cos (2A) + 1}{2} \!$$


 * $$\cos^2\frac{A}{2} = \frac{\cos A + 1}{2} \!$$

Finally we take the square root of both sides
 * $$\cos\frac{A}{2} = \pm\, \sqrt{ \frac{\cos A + 1}{2}} \!$$

4c. Find the value of $$\cos^2(\theta)\!$$ without exponents using the above rules.
 * $$\cos (2\theta) = 2\cos^2 (\theta) - 1\!$$


 * $$2\cos^2 (\theta) - 1 = \cos (2\theta) \!$$


 * $$2\cos^2 (\theta) = \cos (2\theta) + 1\!$$


 * $$\cos^2 (\theta) = \frac{\cos (2\theta) + 1}{2}\!$$

'''4d. (Challenge) Find the value of $$\cos^3(\theta)\!$$ without exponents.''' We begin with our formula
 * $$\cos(A+B)=\cos (A) \cos (B) - \sin (A)\sin (B)\!$$

Following the clue from the previous example we add another A, therefore
 * $$\cos(2\theta+\theta)=\cos (2\theta) \cos (\theta) - \sin (2\theta)\sin (\theta)\!$$

Now expand the expressions
 * $$\cos(2\theta)=(2\cos^2 (\theta) - 1)\!$$


 * $$\sin(2\theta)=2\sin \theta \cos \theta\!$$


 * $$\cos(2\theta+\theta)=((2\cos^2 (\theta) - 1) \cos (\theta)) - ((2\sin \theta \cos \theta)\sin (\theta))\!$$

Clean up the results
 * $$\cos(3\theta)=((2\cos^2 (\theta) - 1) \cos (\theta)) - (2\sin^2 \theta \cos \theta))\!$$


 * $$\cos(3\theta)= \cos \theta((2\cos^2 (\theta) - 1) - (2\sin^2 \theta))\!$$


 * $$\sin^2\theta + \cos^2\theta = 1\!$$


 * $$2\sin^2\theta + 2\cos^2\theta = 2\!$$


 * $$2\sin^2\theta = 2 - 2\cos^2\theta \!$$


 * $$\cos(3\theta)= \cos \theta((2\cos^2 (\theta) - 1) - (2 - 2\cos^2\theta))\!$$


 * $$\cos(3\theta)= \cos \theta(2\cos^2 (\theta) - 1 - 2 + 2\cos^2\theta)\!$$


 * $$\cos(3\theta)= \cos \theta(4\cos^2 (\theta) - 3)\!$$


 * $$\cos(3\theta)= 4\cos^3 (\theta) - 3\cos \theta\!$$


 * $$4\cos^3 (\theta)=\cos(3\theta)+ 3\cos \theta \!$$


 * $$\cos^3 (\theta)=\frac{\cos(3\theta)+ 3\cos \theta}{4} \!$$