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Question: What is the value of \[\cos x – \sin x =\] ?...

What is the value of cos⁡x–sin⁡x=\cos x – \sin x = ?

Explanation

Solution

In this question, we need to find the value of cos⁡x−sin⁡x\cos x-\sin x . The basic trigonometric functions are sine , cosine and tangent. Sine is nothing but a ratio of the opposite side of a right angle to the hypotenuse of the right angle. Similarly, cosine is nothing but a ratio of the adjacent side of a right angle to the hypotenuse of the right angle . Here we need to find the value of cos⁡x−sin⁡x\cos x-\sin x . With the help of the Trigonometric functions , we can find the value of cos⁡x−sin⁡x\cos x-\sin x
Formula used :
1. sin⁡x=cos(π2–x)\sin x = cos\left( \dfrac{\pi}{2} – x \right)
2. cosa–cosb=−2sin(a+b2)sin⁡(a–b2)cosa – cosb = - 2sin\left( \dfrac{a + b}{2} \right)\sin\left( \dfrac{a – b}{2} \right)
Trigonometry table :

Angle0o0^{o}30o30^{o}45o45^{o}60o60^{o}90o90^{o}
Sine0012\dfrac{1}{2}12\dfrac{1}{\sqrt{2}}32\dfrac{\sqrt{3}}{2}11

Complete step by step solution:
Given,
cos⁡x–sin⁡x\cos x – \sin x
We need to find the value of cos⁡x–sin⁡x\cos x – \sin x
By using the identity, sin⁡x=cos(π2–x)\sin x = cos(\dfrac{\pi}{2} – x)
We get ,
cos⁡x–sin⁡x=cos⁡x−cos⁡(π2–x)\cos x – \sin x = \cos x - \cos\left( \dfrac{\pi}{2} – x \right)
From the trigonometry formula,
cosa–cosb=−2sin(a+b2)sin⁡(a–b2)cosa – cosb = - 2sin\left( \dfrac{a + b}{2} \right)\sin\left( \dfrac{a – b}{2} \right)
From comparing the expression cos⁡x−cos⁡(π2–x)\cos x - \cos\left( \dfrac{\pi}{2} – x \right) with the trigonometry formula, a=xa = x and b=(π2–x) b = \left( \dfrac{\pi}{2} – x \right)\
By substituting aa and bb in the formula,
We get,
⇒−2sin(x+(π2)–x2)sin⁡(x−((π2)–x)2) \Rightarrow- 2sin\left( \dfrac{x + \left( \dfrac{\pi}{2} \right) – x}{2} \right)\sin\left( \dfrac{x - \left( \left( \dfrac{\pi}{2} \right) – x \right)}{2} \right)\
By simplifying the term
(x+(π2)–x2)\left( \dfrac{x + \left( \dfrac{\pi}{2} \right) – x}{2} \right)
We get, π4\dfrac{\pi}{4}
Also another term,
(x−((π2)–x)2) \left( \dfrac{x - \left( \left( \dfrac{\pi}{2} \right) – x \right)}{2} \right)\
=(x−(π–2x2)2)= \left( \dfrac{x - \left( \dfrac{\pi – 2x}{2} \right)}{2} \right)
On solving,
We get,
=(2x−(π–2x)22)= \left( \dfrac{\dfrac{2x - \left( \pi – 2x \right)}{2}}{2} \right)
=(2x−(π–2x)2×2)= \left( \dfrac{2x - \left( \pi – 2x \right)}{2 \times 2} \right)
By removing the parentheses,
We get,
=2x−π+2x4=\dfrac{2x - \pi + 2x}{4}
On further simplifying,
We get
=4x−π4=\dfrac{4x - \pi}{4}
On dividing,
We get,
=x−(π4)= x - \left( \dfrac{\pi}{4} \right)
Thus by substituting both the terms,
We get,
=−2sin(π4)sin⁡(x−π4)= - 2sin\left( \dfrac{\pi}{4} \right)\sin\left( x - \dfrac{\pi}{4} \right)
From the trigonometric table, the value of sin⁡(π4)\sin\left( \dfrac{\pi}{4} \right) is 12\dfrac{1}{\sqrt{2}}
By substituting the value ,
We get ,
=−2×12sin⁡(x−π4)= - 2 \times \dfrac{1}{\sqrt{2}}\sin\left( x - \dfrac{\pi}{4} \right)
On simplifying,
We get,
=−2sin⁡(x−π4)= - \sqrt{2}\sin\left( x - \dfrac{\pi}{4} \right)
Thus we get,
cos⁡x–sin⁡x=−2sin⁡(x−π4)\cos x – \sin x = - \sqrt{2}\sin\left( x - \dfrac{\pi}{4} \right)
Final answer :
The value of cos⁡x–sin⁡x=−2sin⁡(x−π4)\cos x – \sin x = - \sqrt{2}\sin\left( x - \dfrac{\pi}{4} \right)

Note: The concept used in this problem is trigonometric identities and ratios. Trigonometric identities are nothing but they involve trigonometric functions including variables and constants. The common technique used in this problem is the use of trigonometric functions . Trigonometric functions are also known as circular functions or geometrical functions.