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If

\begin{equation} \sin(x) + \cos(x) = \frac{7}{5}, \end{equation}

then what's the value of

\begin{equation} \frac{1}{\sin(x)} + \frac{1}{\cos(x)}\text{?} \end{equation}

Meaning the value of $\sin(x)$, $\cos(x)$ (the denominator) without using the identities of trigonometry.

The function $\sin x+\cos x$ could be transformed using some trigonometric identities to a single function. In fact, WolframAlpha says it is equal to $\sqrt2\sin\left(x+\frac\pi4\right)$ and there also are some posts on this site about this equality. So probably in this way we could calculate $x$ from the first equation - and once we know $\sin x$ and $\cos x$, we can calculate $\dfrac{1}{\sin x}+\dfrac{1}{\cos x}$. Is there a simpler solution (perhaps avoiding explicitly finding $x$)?

Rushabh Mehta
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user2161721
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    Note: $ \sin x+ \cos x = \sqrt2\sin(x+\frac{\pi}{4})$ – George Feb 10 '16 at 20:30
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    It is impossible without identities. – N.S.JOHN Feb 26 '16 at 15:37
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    my hint: $(\sin x+\cos x)^2=1+2\sin x\cosx$ – Bhaskara-III Feb 27 '16 at 12:13
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    If this is the sort of post that ends up becoming a HNQ, I think it really, really demeans this site. The entire SE network can just click on the HNQ, and assume this question, and it's answers which are easily found via Wolfram Alpha, represent this sites quality. In this case, not at all. – amWhy Aug 06 '17 at 16:47

10 Answers10

105

Notice, $$\frac{1}{\sin x}+\frac{1}{\cos x}$$ $$=\frac{\sin x+\cos x}{\sin x\cos x}$$ $$=2\cdot \frac{\sin x+\cos x}{2\sin x\cos x}$$ $$=2\cdot \frac{\sin x+\cos x}{(\sin x+\cos x)^2-1}$$ setting the value of $\sin x+\cos x$, $$=2\cdot \frac{\frac 75}{\left(\frac{7}{5}\right)^2-1}$$ $$=\frac{35}{12}$$

Harish Chandra Rajpoot
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19

$$\sin x+\cos x=\frac{7}{5}$$ Let $\sin x=t$ $$\implies t+\sqrt{1-t^2}=\frac{7}{5}$$ Shifting, squaring and simplifying, we get $$25t^2-35t+12=0$$ $$\implies t=\frac{35 \pm 5}{50}$$ Hence, $$\sin x= \frac{4}{5},\ \cos x=\frac{3}{5} \ \text{or} \ \cos x= \frac{4}{5}, \ \sin x=\frac{3}{5}$$ But as we need to find $$\frac{1}{\sin(x)} +\frac{1}{\cos(x)}$$ it becomes $$\frac{5}{4}+\frac{5}{3}=\frac{35}{12}$$

Bhaskara-III
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Nikunj
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18

\begin{align} \sin(x)+\cos(x) &= \frac 75 \\ \left(\sin(x)+\cos(x)\right)^2 &= \left( \frac 75 \right)^2 \\ 1 + 2 \sin(x) \cos(x) &= \frac{49}{25} \\ \sin(x) \cos(x) &= \frac{12}{25} \end{align}

\begin{align} \frac{1}{\sin(x)} + \frac{1}{\cos(x)} &= \frac{\sin(x) + \cos(x)}{\sin(x) \cos(x)}\\ &= \frac{\left(\frac{7}{5}\right)}{\left(\frac{12}{25}\right)}\\ &= \frac{7 \times 25}{5\times 12}\\ &= \frac{35}{12} \end{align}

Steven Alexis Gregory
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5

$$ \begin{align} \sin(x)+\cos(x)=\frac75 &\implies1-\sin^2(x)=\frac{49}{25}-\frac{14}5\sin(x)+\sin^2(x)\\ &\implies\sin^2(x)-\frac75\sin(x)+\frac{12}{25}=0\\ &\implies\sin(x)=\frac{7\pm1}{10}\\ &\implies\sin(x)\in\left\{\frac35,\frac45\right\} \end{align} $$ Since $\cos(x)=\frac75-\sin(x)\gt0$, if $\sin(x)=\frac35$ then $\cos(x)=\frac45$ and vice-versa. Therefore, $$ \frac1{\sin(x)}+\frac1{\cos(x)}=\frac{35}{12} $$

robjohn
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4

$$ s + c = \frac{7}{5};\quad s^2 + c^2 \frac{35}{12} = 1; $$

Solving quadratic equation by elimination of one of the two variables

$$ s= \left(\frac{4}{5}, \frac{3}{5} \right );\quad c = \left(\frac{3}{5}, \frac{4}{5} \right) ; $$

respectively. They are interchangeable. So only one result is obtained with either of two inputs:

$$ \frac{1}{s} + \frac{1}{c} =\frac{5}{4} +\frac{5}{3} = \frac{35}{12}. $$

Narasimham
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3

If

$$\sin x+\cos x=a\qquad\text{and}\qquad{\frac{1}{\sin x}}+{\frac{1}{\cos x}}=b$$ then

$$a=b\sin x\cos x\qquad\text{and}\qquad a^2=1+2\sin x\cos x=1+\frac{2a}{b}$$

from which it follows that

$$b={\frac{2a}{a^2-1}}$$

Letting $a=\frac{7}{5}$, we have

$$b={\frac{\frac{14}{5}}{\frac{49}{25}-1}}={\frac{70}{24}}={\frac{35}{12}}$$

Bhaskara-III
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Barry Cipra
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3

Assume that,

$\sin x=a, \cos x=b $

Given that : $$\sin x+\cos x=\frac75$$ $$a+b=\frac75\tag 1$$ $$\sin^2 x+\cos^2 x=1$$ $$a^2 +b^2 =1$$ $$(a+b)^2-2ab =1$$ $$(7/5)^2-2ab =1$$ $$ab=12/25\tag 2$$ solving (1), (2), i get $a=3/5, b=4/5$ therefore, $$\frac1{\sin x}+\frac1{\cos x}=\frac1a+\frac1b$$$$=\frac{1}{3/5}+\frac{1}{4/5}$$ $$=\frac{35}{12}$$

jeanne clement
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2

initially i have

$\sin x+\cos x={7\over5}$

took the squares,

$\sin^2x+\cos^2x+2\sin x\cos x={49\over 25}$

$1+2\sin x\cos x={49\over 25}$

$\sin x\cos x={49\over 50}-{1\over 2}={12\over 25}$

$$\frac{1}{\sin x}+\frac{1}{\cos x}$$ $$\frac{\sin x+\cos x}{\sin x\cos x}$$ $$\frac{{7\over 5}}{{12\over 25}}$$ $${35\over 12}$$

1

$$\sin x+\cos x=\frac75\tag1$$ take the squares, $$(\sin x+\cos x)^2=\left(\dfrac75\right)^2$$ $$\sin x\cos x=\frac{12}{25}\tag2$$ $$\therefore \dfrac{1}{\sin x}+\dfrac{1}{\cos x}$$ $$=\dfrac{\sin x+\cos x}{\sin x\cos x}$$ $$=\dfrac{\left(\dfrac{7}{5}\right)}{\left(\dfrac{12}{25}\right)}$$ $$=\dfrac{35}{12}$$ Hope it helps!

1

$$ x+y= p\tag1$$ Square, since $( x^2+y^2=1 )$ $$ 1+ 2 x\;y = p^2, \; x y= \dfrac{p^2-1}{2} \tag2$$ From (1) and (2) $$ \dfrac{1}{x}+ \dfrac{1}{y} = \dfrac{x+y}{x y}= \dfrac{2p}{p^2-1} $$ $$ = \dfrac{35}{12},\;$$ if $$\;p= \dfrac{7}{5} $$

Narasimham
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