Question 13, Exercise 10.1

Solutions of Question 13 of Exercise 10.1 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Express each of the following in the form $r\,\,\sin \left( \theta +\phi \right)$ where terminal ray of $\theta$ and $\phi$ are in the first quadrant $4\sin \theta +3\cos \theta .$ Identifying $4\sin \theta +3\cos \theta$ with $r\sin(\theta + \varphi)$ gives $$4\sin \theta +3\cos \theta=r\cos\varphi\sin\theta+r\sin\varphi\cos\theta --- (1)$$ So $4=r\cos\varphi$ and $3=r\sin\varphi$.

Squaring and adding $$\begin{align} &r^2\cos^2 \varphi+r^2\sin^2\varphi = 4^2+3^2\\ \implies &r^2\left(\cos^2\varphi+\sin^2\varphi\right) = 16+9\\ \implies &r^2 = 25\\ \implies &r=5 \end{align}$$ Also $$\begin{align}r\cos\varphi = 4 \implies 5\cos\varphi = 4 \implies \cos\varphi =\dfrac{4}{5}\end{align}$$ and $$\begin{align}r\sin\varphi = 3 \implies 5\sin\varphi = 3 \implies \sin\varphi =\dfrac{3}{5}.\end{align}$$ Thus, from (1), we have $$\begin{align}&4\sin\theta +3\cos\theta \\ &=5 \left(\dfrac{4}{5}\sin\theta +\dfrac{3}{5}\cos\theta\right)\\ &=r\left(\cos\varphi\sin\theta +\sin\varphi\cos\theta\right)\\ &=r\sin(\theta+\varphi),\\ &\text{ where } \sin\varphi=\dfrac{3}{5}, \cos\varphi=\dfrac{4}{5} \text{ and } r=5.\end{align}$$

Express each of the following in the form $r\,\,\sin \left( \theta +\phi \right)$ where terminal ray of $\theta$ and $\phi$ are in the first quadrant $15\sin \theta +8\cos \theta .$

Identifying $15\sin \theta +8\cos \theta$ with $r\sin(\theta + \varphi)$ gives $$15\sin \theta +8\cos \theta=r\cos\varphi\sin\theta+r\sin\varphi\cos\theta --- (1)$$ So $15=r\cos\varphi$ and $8=r\sin\varphi$.

Squaring and adding $$\begin{align} &r^2\cos^2 \varphi+r^2\sin^2\varphi = 15^2+8^2\\ \implies &r^2\left(\cos^2\varphi+\sin^2\varphi\right) = 225+64\\ \implies &r^2 = 289\\ \implies &r=17 \end{align}$$ Also $$\begin{align}r\cos\varphi = 15 \implies 17\cos\varphi = 15 \implies \cos\varphi =\dfrac{15}{17}\end{align}$$ and $$\begin{align}r\sin\varphi = 8 \implies 17\sin\varphi = 8 \implies \sin\varphi =\dfrac{8}{17}.\end{align}$$ Thus, from (1), we have $$\begin{align}&15\sin\theta +8\cos\theta \\ &=17 \left(\dfrac{15}{17}\sin\theta +\dfrac{8}{17}\cos\theta\right)\\ &=r\left(\cos\varphi\sin\theta +\sin\varphi\cos\theta\right)\\ &=r\sin(\theta+\varphi),\\ &\text{ where } \sin\varphi=\dfrac{8}{17}, \cos\varphi=\dfrac{15}{17} \text{ and } r=17.\end{align}$$

Express each of the following in the form $r\,\,\sin \left( \theta +\phi \right)$ where terminal ray of $\theta$ and $\phi$ are in the first quadrant $2\sin \theta -5\cos \theta .$ Identifying $2\sin \theta -5\cos \theta$ with $r\sin(\theta + \varphi)$ gives $$2\sin \theta +-5\cos \theta=r\cos\varphi\sin\theta+r\sin\varphi\cos\theta --- (1)$$ So $2=r\cos\varphi$ and $-5=r\sin\varphi$.

Squaring and adding $$\begin{align} &r^2\cos^2 \varphi+r^2\sin^2\varphi = 2^2+(-5)^2\\ \implies &r^2\left(\cos^2\varphi+\sin^2\varphi\right) = 4+25\\ \implies &r^2 = 29\\ \implies &r=\sqrt{29} \end{align}$$ Also $$\begin{align}r\cos\varphi = 2 \implies \sqrt{29}\cos\varphi = 2 \implies \cos\varphi =\dfrac{2}{\sqrt{29}}\end{align}$$ and $$\begin{align}r\sin\varphi = -5 \implies \sqrt{29}\sin\varphi = -5 \implies \sin\varphi =\dfrac{-5}{\sqrt{29}}.\end{align}$$ Thus, from (1), we have $$\begin{align}&2\sin\theta -5\cos\theta \\ &=\sqrt{29} \left(\dfrac{2}{\sqrt{29}}\sin\theta +\dfrac{-5}{\sqrt{29}}\cos\theta\right)\\ &=r\left(\cos\varphi\sin\theta +\sin\varphi\cos\theta\right)\\ &=r\sin(\theta+\varphi),\\ &\text{ where } \sin\varphi=\dfrac{-5}{\sqrt{29}}, \cos\varphi=\dfrac{2}{\sqrt{29}} \text{ and } r=\sqrt{29}.\end{align}$$

Express each of the following in the form $r\,\,\sin \left( \theta +\phi \right)$ where terminal ray of $\theta$ and $\phi$ are in the first quadrant $\sin \theta +\cos \theta .$ Identifying $\sin\theta+\cos\theta$ with $r\sin(\theta + \varphi)$ gives $$\sin\theta+\cos\theta=r\cos\varphi\sin\theta+r\sin\varphi\cos\theta --- (1)$$ So $1=r\cos\varphi$ and $1=r\sin\varphi$.

Squaring and adding $$\begin{align} &r^2\cos^2 \varphi+r^2\sin^2\varphi = 1^2+1^2\\ \implies &r^2\left(\cos^2\varphi+\sin^2\varphi\right) = 1+1\\ \implies &r^2 = 2\\ \implies &r=\sqrt{2} \end{align}$$ Also $$\begin{align}r\cos\varphi = 1 \implies \sqrt{2}\cos\varphi = 1 \implies \cos\varphi =\dfrac{1}{\sqrt{2}}\end{align}$$ and $$\begin{align}r\sin\varphi = 1 \implies \sqrt{2}\sin\varphi = 1 \implies \sin\varphi =\dfrac{1}{\sqrt{2}}.\end{align}$$ Thus, from (1), we have $$\begin{align}&\sin\theta+\cos\theta \\ &=\sqrt{2} \left(\dfrac{1}{\sqrt{2}}\sin\theta +\dfrac{1}{\sqrt{2}}\cos\theta\right)\\ &=r\left(\cos\varphi\sin\theta +\sin\varphi\cos\theta\right)\\ &=r\sin(\theta+\varphi),\\ &\text{ where } \sin\varphi=\dfrac{1}{\sqrt{2}}, \cos\varphi=\dfrac{1}{\sqrt{2}} \text{ and } r=\sqrt{2}.\end{align}$$

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