4.3 Contents

1. Right Angle Identities

2. Law of Sines

3. Law of Cosines

4. Sum & Difference Identities

5. Product-to-Sum Identities

6. Sum-to-Product Identities

7. Double Angle Identities

8. Half Angle Identities

9. Square Identities

1. Functions of Inverses 🔧

4.3.1 Right Angle Identities

$$\sin^2(\theta)+\cos^2(\theta)=1$$ $$\tan^2(\theta)+1=\sec^2(\theta)$$ $$\cot^2(\theta)+1=\csc^2(\theta)$$

Proofs

In each case, given the right angle theorem, divide by the hypotenuse, $x$ component, and $y$ component respectively $$\frac{h^2=x^2+y^2}{h^2}\medspace\to\medspace 1=\frac{x^2}{h^2}+\frac{y^2}{h^2}$$ $$\frac{h^2=x^2+y^2}{x^2}\medspace\to\medspace \frac{h^2}{x^2}=1+\frac{y^2}{x^2}$$ $$\frac{h^2=x^2+y^2}{y^2}\medspace\to\medspace \frac{h^2}{y^2}=\frac{x^2}{y^2}+1$$ In each case, substitute the right angle definition

4.3.2 Law of Sines

$$\frac{\sin(A)}{a}=\frac{\sin(B)}{b}=\frac{\sin(C)}{c}$$

Proof

Given the triangle area, substitute the height orthogonal to each base with its right angle definition for sine $$\frac{1}{2}\cdot b\cdot c\cdot\sin(A)=\frac{1}{2}\cdot c\cdot a\cdot\sin(B)=\frac{1}{2}\cdot a\cdot b\cdot\sin(C)$$ Multiply by $2/(a\cdot b\cdot c)$

4.3.3 Law of Cosines

$$c^2=a^2+b^2-2\cdot a\cdot b\cdot\cos(C)$$

Proof

Insert an orthogonal line to $c$ to represent the height of the triangle, dividing $c$ as $c_1$ and $c_2$ Find $c_1$ and $c_2$ by taking the cosine of $A$ and $B$ $$\cos(A)=\frac{c_2}{b}\qquad\cos(B)=\frac{c_1}{a}$$ Solve for $c_1$ and $c_2$ $$c_2=b\cdot\cos(A)\qquad c_1=a\cdot\cos(B)$$ Add $c_1$ to $c_2$ for $c$ $$c_2+c_1=c=b\cdot\cos(A)+a\cdot\cos(B)$$ Multiply by $c$ $$c^2=b\cdot c\cdot\cos(A)+a\cdot c\cdot\cos(B)$$ Repeat the process for the other two sides $$a^2=c\cdot a\cdot\cos(B)+b\cdot a\cdot\cos(C)$$ $$b^2=a\cdot b\cdot\cos(C)+c\cdot b\cdot\cos(A)$$ Add the last two equations and subtract the first $$a^2+b^2-c^2=c\cdot a\cdot\cos(B)+b\cdot a\cdot\cos(C)+a\cdot b\cdot\cos(C)+c\cdot b\cdot\cos(A)-b\cdot c\cdot\cos(A)-a\cdot c\cdot\cos(B)$$ Simplify $$a^2+b^2-c^2=2\cdot a\cdot b\cdot\cos(C)$$ Isolate $c^2$

4.3.4 Sum & Difference Identities

$$\sin(a\pm b)=\sin(a)\cdot\cos(b)\pm\cos(a)\cdot\sin(b)$$ $$\cos(a\pm b)=\cos(a)\cdot\cos(b)\mp\sin(a)\cdot\sin(b)$$ $$\tan(a\pm b)=\frac{\tan(a)\pm\tan(b)}{1\mp\tan(a)\cdot\tan(b)}$$ $$\cot(a\pm b)=\frac{\cot(a)\cdot\cot(b)\mp1}{\cot(b)\pm\cot(a)}$$

Proof of Sine & Cosine

On the unit circle, plot points for $(1,0)$, and for the angles $b$, $a-b$, and $a$ respectively in a positive rotation Determine the coordinates of each point given their angles using the unit circle definitions The distances displayed are equal to each other. Use the distance formula for each set equal to each other. $$\sqrt{\big(1-\cos(a-b)^2\big)^2+\big(0-\sin(a-b)^2\big)^2}=\sqrt{\big(\cos(b)-\cos(a)\big)^2+\big(\sin(b)-\sin(a)\big)^2}$$ Square $$\big(1-\cos(a-b)^2\big)^2+\big(0-\sin(a-b)^2\big)^2=\big(\cos(b)-\cos(a)\big)^2+\big(\sin(b)-\sin(a)\big)^2$$ Expand the left $$1-2\cdot\cos(a-b)+\cos^2(a-b)+\sin^2(a-b)=\big(\cos(b)-\cos(a)\big)^2+\big(\sin(b)-\sin(a)\big)^2$$ Use the right angle identity to simplify to one $$1-2\cdot\cos(a-b)+1=\big(\cos(b)-\cos(a)\big)^2+\big(\sin(b)-\sin(a)\big)^2$$ Expand the right $$2-2\cdot\cos(a-b)=\cos^2(b)-2\cdot\cos(a)\cdot\cos(b)+\cos^2(a)+\sin^2(b)-2\cdot\sin(a)\cdot\sin(b)+\sin^2(a)$$ Use the right angle identity to simplify to 1 in two instances $$2-2\cdot\cos(a-b)=-2\cdot\cos(a)\cdot\cos(b)-2\cdot\sin(a)\cdot\sin(b)+2$$ Subtract 2 $$-2\cdot\cos(a-b)=-2\cdot\cos(a)\cdot\cos(b)-2\cdot\sin(a)\cdot\sin(b)$$ Divide by $-2$ $$\cos(a-b)=\cos(a)\cdot\cos(b)+\sin(a)\cdot\sin(b)$$ Substitute $-b$ for $b$ $$\cos\big(a-(-b)\big)=\cos(a)\cdot\cos(-b)+\sin(a)\cdot\sin(-b)$$ Apply the even/odd identities $$\cos(a+b)=\cos(a)\cdot\cos(b)-\sin(a)\cdot\sin(b)$$ Substitute $\pi/2-(a+b)$ for $a+b$ and $\pi/2-a$ for $a$ $$\cos\bigg(\frac{\pi}{2}-a-b\bigg)=\cos\bigg(\frac{\pi}{2}-a\bigg)\cdot\cos(b)-\sin\bigg(\frac{\pi}{2}-a\bigg)\cdot\sin(b)$$ Apply the complimentary angle identities $$\sin(a-b)=\sin(a)\cdot\cos(b)-\cos(a)\cdot\sin(b)$$ Substitute $-b$ for $b$ $$\sin\big(a-(-b)\big)=\sin(a)\cdot\cos(-b)-\cos(a)\cdot\sin(-b)$$ Apply the even/odd identities $$\sin(a+b)=\sin(a)\cdot\cos(b)+\cos(a)\cdot\sin(b)$$

Proof of Tangent

Substitute tangent for its sine/cosine cofunction $$\tan(a\pm b)=\frac{\sin(a\pm b)}{\cos(a\pm b)}$$ Substitute the sum & difference identities for sine and cosine $$\tan(a\pm b)=\frac{\sin(a)\cdot\cos(b)\pm\cos(a)\cdot\sin(b)}{\cos(a)\cdot\cos(b)\mp\sin(a)\cdot\sin(b)}$$ Divide the numerator and denominator by $\cos(a)\cdot\cos(b)$ $$\tan(a\pm b)= \frac{\big(\sin(a)\cdot\cos(b)\pm\cos(a)\cdot\sin(b)\big)/\big(\cos(a)\cdot\cos(b)\big)} {\big(\cos(a)\cdot\cos(b)\mp\sin(a)\cdot\sin(b)\big)/\big(\cos(a)\cdot\cos(b)\big)}$$ Cancel like terms $$\tan(a\pm b)=\frac{\sin(a)/\cos(a)\pm\sin(b)/\cos(b)}{1\mp\big(\sin(a)\cdot\sin(b)\big)/\big(\cos(a)\cdot\cos(b)\big)}$$ Substitute the tangent cofunction for the sines/cosines

Proof of Cotangent

Substitute cotangent for its cosine/sine cofunction $$\cot(a\pm b)=\frac{\cos(a\pm b)}{\sin(a\pm b)}$$ Substitute the sum & difference identities for cosine and sine $$\cot(a\pm b)=\frac{\cos(a)\cdot\cos(b)\mp\sin(a)\cdot\sin(b)}{\sin(a)\cdot\cos(b)\pm\cos(a)\cdot\sin(b)}$$ Divide the numerator and denominator by $\sin(a)\cdot\sin(b)$ $$\cot(a\pm b)= \frac{\big(\cos(a)\cdot\cos(b)\mp\sin(a)\cdot\sin(b)\big)/\big(\sin(a)\cdot\sin(b)\big)} {\big(\sin(a)\cdot\cos(b)\pm\cos(a)\cdot\sin(b)\big)/\big(\sin(a)\cdot\sin(b)\big)}$$ Cancel like terms $$\cot(a\pm b)=\frac{\big(\cos(a)\cdot\cos(b)\big)/\big(\sin(a)\cdot\sin(b)\big)\mp 1}{\cos(b)/\sin(b)\pm\cos(a)/\sin(a)}$$ Substitute the cotangent cofunction for the cosines over sines

4.3.5 Product-to-Sum Identities

$$2\cdot\sin(a)\cdot\sin(b)=\cos(a-b)-\cos(a+b)$$ $$2\cdot\cos(a)\cdot\cos(b)=\cos(a-b)+\cos(a+b)$$ $$2\cdot\sin(a)\cdot\cos(b)=\sin(a+b)-\sin(a-b)$$ $$2\cdot\cos(a)\cdot\sin(b)=\sin(a+b)-\sin(a-b)$$

Proofs

In each case, substitute the sum and difference identities for the terms on the right, and simplify

4.3.6 Sum-to-Product Identities

$$\sin(a)\pm\sin(b)=2\cdot\sin\bigg(\frac{a \pm b}{2}\bigg)\cdot\cos\bigg(\frac{a\mp b}{2}\bigg)$$ $$\cos(a)+\cos(b)=2\cdot\cos\bigg(\frac{a+b}{2}\bigg)\cdot\cos\bigg(\frac{a-b}{2}\bigg)$$ $$\cos(a)-\cos(b)=-2\cdot\sin\bigg(\frac{a+b}{2}\bigg)\cdot\sin\bigg(\frac{a-b}{2}\bigg)$$

Proofs

Using the angles from the product-to-sum identities, let $a=(u+v)/2$ and $b=(u-v)/2$ so that $$a+b=\frac{u+v}{2}+\frac{u-v}{2}=u$$ $$a-b=\frac{u+v}{2}-\frac{u-v}{2}=v$$ Substitute the terms with $a$ and $b$ for the terms with $u$ and $v$ into the product-to-sum identities

4.3.7 Double Angle Identities

$$\sin(2\cdot\theta)=2\cdot\sin(\theta)\cdot\cos(\theta)$$ $$\csc(2\cdot\theta)=\frac{1}{2}\cdot\sec(\theta)\cdot\csc(\theta)$$ $$\cos(2\cdot\theta)=\cos^2(\theta)-\sin^2(\theta)$$ $$\sec(2\cdot\theta)=\frac{1}{\cos^2(\theta)-\sin^2(\theta)}$$ $$\tan(2\cdot\theta)=\frac{2\cdot\tan(\theta)}{1-\tan^2(\theta)}$$ $$\cot(2\cdot\theta)=\frac{\cot^2(\theta)-1}{2\cdot\cot(\theta)}$$

Proofs

In each case, start with the sum identity for the corresponding function using the same angle for both variables. For secant and cosecant, use $\cos(a\pm b)^{-1}$ and $\sin(a\pm b)^{-1}$ respectively.

Example

The circle segment area expression can be simplified by using the sine double angle identity $$\frac{r^2}{2}\cdot\Big(\theta-2\cdot\sin\Big(\frac{\theta}{2}\Big)\cdot\cos\Big(\frac{\theta}{2}\Big)\Big)$$ $$\therefore\frac{r^2}{2}\cdot\big(\theta-\sin(\theta)\big)$$

4.3.8 Half Angle Identities

$$\sin\bigg(\frac{\theta}{2}\bigg)=\pm\sqrt{\frac{1-\cos(\theta)}{2}}$$ $$\cos\bigg(\frac{\theta}{2}\bigg)=\pm\sqrt{\frac{1+\cos(\theta)}{2}}$$ $$\tan\bigg(\frac{\theta}{2}\bigg)=\pm\sqrt{\frac{1-\cos(\theta)}{1+\cos(\theta)}}$$ $$\cot\bigg(\frac{\theta}{2}\bigg)=\pm\sqrt{\frac{1+\cos(\theta)}{1-\cos(\theta)}}$$ In each case, only one solution exists, and signs must be determined by the quadrant of $\theta/2$

Proof of Sine

Given the cosine double angle identity, use $\theta$ for $2\cdot\theta$ and $\theta/2$ for $\theta$ $$\cos(\theta)=\cos^2\bigg(\frac{\theta}{2}\bigg)-\sin^2\bigg(\frac{\theta}{2}\bigg)$$ Substitute the right angle identity for $\cos^2(\theta)$ $$\cos(\theta)=1-2\cdot\sin^2\bigg(\frac{\theta}{2}\bigg)$$ Subtract 1 $$\cos(\theta)-1=-2\cdot\sin^2\bigg(\frac{\theta}{2}\bigg)$$ Divide by –2 $$\frac{1-\cos(\theta)}{2}=\sin^2\bigg(\frac{\theta}{2}\bigg)$$ Take the square root

Proof of Cosine

Given the cosine double angle identity, use $\theta$ for $2\cdot\theta$ and $\theta/2$ for $\theta$ $$\cos(\theta)=\cos^2\bigg(\frac{\theta}{2}\bigg)-\sin^2\bigg(\frac{\theta}{2}\bigg)$$ Substitute the right angle identity for $\sin^2(\theta)$ $$\cos(\theta)=2\cdot\cos^2\bigg(\frac{\theta}{2}\bigg)-1$$ Add 1 $$1+\cos(\theta)=2\cdot\cos^2\bigg(\frac{\theta}{2}\bigg)$$ Divide by 2 $$\frac{1+\cos(\theta)}{2}=\cos^2\bigg(\frac{\theta}{2}\bigg)$$ Take the square root

Proof of Tangent

Divide the sine half angle identity by the cosine half angle identity $$\bigg|\sin\bigg(\frac{\theta}{2}\bigg)\bigg|\bigg/\bigg|\cos\bigg(\frac{\theta}{2}\bigg)\bigg|=\sqrt{\frac{1-\cos(\theta)}{2}}\bigg/\sqrt{\frac{1+\cos(\theta)}{2}}$$ Substitute sine/cosine for its tangent cofunction $$\bigg|\tan\bigg(\frac{\theta}{2}\bigg)\bigg|=\sqrt{\frac{1-\cos(\theta)}{2}}\bigg/\sqrt{\frac{1+\cos(\theta)}{2}}$$ Factor the root $$\bigg|\tan\bigg(\frac{\theta}{2}\bigg)\bigg|=\sqrt{\frac{1-\cos(\theta)}{2}\bigg/\frac{1+\cos(\theta)}{2}}$$ Simplify

Proof of Cotangent

Divide the cosine half angle identity by the sine half angle identity $$\bigg|\cos\bigg(\frac{\theta}{2}\bigg)\bigg|\bigg/\bigg|\sin\bigg(\frac{\theta}{2}\bigg)\bigg|=\sqrt{\frac{1+\cos(\theta)}{2}}\bigg/\sqrt{\frac{1-\cos(\theta)}{2}}$$ Substitute cosine/sine for its cotangent cofunction $$\bigg|\cot\bigg(\frac{\theta}{2}\bigg)\bigg|=\sqrt{\frac{1+\cos(\theta)}{2}}\bigg/\sqrt{\frac{1-\cos(\theta)}{2}}$$ Factor the root $$\bigg|\cot\bigg(\frac{\theta}{2}\bigg)\bigg|=\sqrt{\frac{1+\cos(\theta)}{2}\bigg/\frac{1-\cos(\theta)}{2}}$$ Simplify

4.3.9 Square Identities

$$\sin^2(\theta)=\frac{1-\cos(2\cdot\theta)}{2}$$ $$\cos^2(\theta)=\frac{1+\cos(2\cdot\theta)}{2}$$ $$\tan^2(\theta)=\frac{1-\cos(2\cdot\theta)}{1+\cos(2\cdot\theta)}$$ $$\cot^2(\theta)=\frac{1+\cos(2\cdot\theta)}{1-\cos(2\cdot\theta)}$$

Proofs

In each case, start with the half angle identity for the corresponding function, use $\theta$ for $2\cdot\theta$ and $\theta/2$ for $\theta$, and square the equations

4.3.A Functions of Inverses 🔧

$$\tan\big(\cos^{-1}(x)\big)=\frac{\sqrt{1-x^2}}{x}$$ $$\sin\big(\cos^{-1}(x)\big)=\sqrt{1-x^2}$$ $$\sin\big(\tan^{-1}(x)\big)=\frac{x}{\sqrt{x^2+1}}$$ $$\cos\big(\tan^{-1}(x)\big)=\frac{1}{\sqrt{x^2+1}}$$ $$\cos\big(\sin^{-1}(x)\big)=\sqrt{1-x^2}$$ $$\tan\big(\sin^{-1}(x)\big)=\frac{x}{\sqrt{1-x^2}}$$

Proof of $\tan\big(\cos^{-1}(x)\big)$ and $\sin\big(\cos^{-1}(x)\big)$

Proof of $\sin\big(\tan^{-1}(x)\big)$ and $\cos\big(\tan^{-1}(x)\big)$

Proof of $\cos\big(\sin^{-1}(x)\big)$ and $\tan\big(\sin^{-1}(x)\big)$