Tag Archives: trigonometric identities

October 9, 2025 · 6:37 am

Trigonometric Exact Values

Find exactly sin(18^\circ)

We must be able to find an arithmetic combination of the exact values we knew to find 18.

    \begin{equation*}90=5\times 18\end{equation}

    \begin{equation*}90-3(18)=2(18)\end{equation}

I re-arranged as above, so I could take advantage of cos(90)=0 and sin(90)=1

Useful identities
sin(2x)=2sin(x)cos(x)
cos(2x)=cos^2(x)-sin^2(x)=1-2sin^2(x)=2cos^2(x)-1
sin(3x)=3sin(x)-4sin^3(x)
cos(3x)=4cos^3(x)-3cos(x)
sin(A-B)=sin(A)cos(B)-sin(B)cos(A)
cos^2(x)=1-sin^2(x)

    \begin{equation*}sin(90-3(18))=sin(2(18))\end{equation}

    \begin{equation*}sin(90)cos(3(18))-sin(3(18))cos(90)=2sin(18)cos(18)\end{equation}

    \begin{equation*}cos(3(18))=2sin(18)cos(18)\end{equation}

    \begin{equation*}4cos^3(18)-3cos(18)=2sin(18)cos(18)\end{equation}

    \begin{equation*}4cos^3(18)-3cos(18)-2sin(18)cos(18)=0\end{equation}

    \begin{equation*}cos(18)(4cos^2(18)-3-2sin(18))=0\end{equation}

Hence,

    \begin{equation*}4cos^2(18)-2sin(18)-3=0\end{equation}

    \begin{equation*}4-4sin^2(18)-2sin(18)-3=0\end{equation}

    \begin{equation*}-4sin^2(18)-2sin(18)+1=0\end{equation}

Use the quadratic equation formula

    \begin{equation*}sin(18)=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\end{equation}

    \begin{equation*}sin(18)=\frac{2 \pm \sqrt{4-4(-4)(1)}}{-8}\end{equation}

    \begin{equation*}sin(18)=\frac{2 \pm \sqrt{20}}{-8}\end{equation}

    \begin{equation*}sin(18)=\frac{-2 \mp 2\sqrt{5}}{8}\end{equation}

    \begin{equation*}sin(18)=\frac{-1 \mp \sqrt{5}}{4}\end{equation}

As sin(18)>0, sin(18)=\frac{-1+\sqrt{5}}{4}

sin(18)=\frac{-1+\sqrt{5}}{4}

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Filed under Addition and Subtraction Identities, Algebra, Identities, Quadratic, Quadratics, Solving, Solving Equations, Solving Trig Equations, Trigonometry, Year 11 Specialist Mathematics

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February 3, 2025 · 9:11 am

Differentiating Trigonometric Functions

In the last post we looked at two trig limits:

(1)   \begin{equation*}\lim_{x \to 0}\frac{sin(x)}{x}=1\end{equation*}

(2)   \begin{equation*}\lim_{x \to 0}\frac{1-cos(x)}{x}=0\end{equation*}

We are going to use these two limits to differentiate sine and cosine functions from first principals.

    \begin{equation*}f(x)=sin(x)\end{equation}

    \begin{equation*}f'(x)=\lim\limits_{h \to 0}\frac{sin(x+h)-sin(x)}{h}\end{equation}

Use the trig identity

    \begin{equation*}sin(A+B)=sinAcosB+sinBcosA\end{equation}

    \begin{equation*}f'(x)=\lim\limits_{h \to 0}\frac{sin(x)cos(h)+sin(h)cos(x)-sin(x)}{h}\end{equation}

    \begin{equation*}f'(x)=\lim\limits_{h \to 0}(\frac{sin(x)(cos(h)-1)}{h}+\frac{sin(h)cos(x)}{h})\end{equation}

    \begin{equation*}f'(x)=sin(x)\lim\limits_{h \to 0}(\frac{(cos(h)-1)}{h}+cos(x)\lim\limits_{h \to 0}\frac{sin(h)}{h}\end{equation}

    \begin{equation*}f'(x)=sin(x)\lim\limits_{h \to 0}(\frac{-(-cos(h)+1)}{h}+cos(x)\lim\limits_{h \to 0}\frac{sin(h)}{h}\end{equation}

Evaluate the limits

    \begin{equation*}f'(x)=sin(x)\times 0+cos(x)\times (1)=cos(x)\end{equation}

Hence, \frac{d}{dx}sin(x)=cos(x).

Now we are going to do the same for f(x)=cos(x).

    \begin{equation*}f'(x)=\lim\limits_{h \to 0}\frac{cos(x+h)-cos(x)}{h}\end{equation}

Use the trigonometric identity

    \begin{equation*}cos(A+B)=cosAcosB-sinAsinB\end{equation}

    \begin{equation*}f'(x)=\lim\limits_{h \to 0}\frac{cos(x)cos(h)-sin(x)sin(h)-cos(x)}{h}\end{equation}

    \begin{equation*}f'(x)=\lim\limits_{h \to 0}\frac{cos(x)(cos(h)-1)-sin(x)sin(h)}{h}\end{equation}

    \begin{equation*}f'(x)=cos(x)\lim\limits_{h \to 0}\frac{-(1-cos(h))}{h}-sin(x)\lim\limits_{h \to 0}\frac{sin(h)}{h}\end{equation}

Evaluate the limits

    \begin{equation*}f'(x)=cos(x)\times(0)-sin(x)\times (1)=-sin(x)\end{equation}

Hence \frac{d}{dx} cos(x)=-sin(x)

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December 16, 2024 · 9:19 am

Interesting Sum

S=\sum_{n=1}^\infty (tan^{-1}(\frac{2}{n^2})), find S.

I came across this sum in An Imaginary Tale by Nahin and I was fascinated.

Let tan(\alpha)=n+1 and tan(\beta)=n-1.

Remember
tan(\alpha-\beta)=\frac{tan(\alpha)-tan(\beta)}{1+tan(\alpha)tan(\beta)}
Hence,
tan(\alpha-\beta)=\frac{(n+1)-(n-1)}{1+(n+1)(n-1)}
tan(\alpha-\beta)=\frac{2}{1+n^2-1}
tan(\alpha-\beta)=\frac{2}{n^2}
Therefore,
\alpha-\beta=tan^{-1}(\frac{2}{n^2})
and
\alpha=tan^{-1}(n+1) and \beta=tan^{-1}(n-1)

tan^{-1}(n+1)-tan^{-1}(n-1)=tan^{-1}(\frac{2}{n^2})

Which means,

    \begin{equation*}S=\sum_{n=1}^\infty(tan^{-1}(n+1)-tan^{-1}(n-1))\end{equation}

Let’s try a few partial sums

S_4=tan^{-1}(2)-tan^{-1}(0)+tan^{-1}(3)-tan^{-1}(1)+tan^{-1}(4)-tan^{-1}(2)+tan^{-1}(5)-tan^{-1}(3)

S_4=-tan^{-1}(0)+-tan^{-1}(1)+tan^{-1}(4)+tan^{-1}(5)

S_6=tan^{-1}(2)-tan^{-1}(0)+tan^{-1}(3)-tan^{-1}(1)+tan^{-1}(4)-tan^{-1}(2)+tan^{-1}(5)-tan^{-1}(3)+tan^{-1}(6)-tan^{-1}(4)+tan^{-1}(7)-tan^{-1}(5)

S_6=-tan^{-1}(0)+-tan^{-1}(1)+tan^{-1}(6)+tan^{-1}(7)

Hence, S_N=-tan^{-1}(0)+-tan^{-1}(1)+tan^{-1}(N)+tan^{-1}(N+1)

S_N=-\frac{\pi}{4}-0+tan^{-1}(N)+tan^{-1}(N+1)

What happens as N\rightarrow \infty ?

\lim\limits_{N\to \infty}\ S_N=-\frac{\pi}{4}+\frac{\pi}{2}+\frac{\pi}{2}=\frac{3\pi}{4}

Because we know tan(\frac{\pi}{2}) is undefined.

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