Question

prove that : √2cos((π/4)+x) = cosx - sinx​

Answer 1

$\large\underline{\sf{Solution-}}$

Consider,

$\rm :\longmapsto\: \sqrt{2} \: cos\bigg[\dfrac{\pi}{4} + x \bigg] \:$

We know,

$\boxed{ \tt{ \: cos(x + y) = cosx \: cosy \: - \: sinx \: siny \: }}$

So, using this

$\rm = \: \sqrt{2}\bigg[cosx \: cos\dfrac{\pi}{4} - sinx \: sin\dfrac{\pi}{4}\bigg]$

We know,

$\: \: \: \: \: \boxed{ \tt{ \: cos\dfrac{\pi}{4} = \frac{1}{ \sqrt{2} } \: }} \: and \: \: \boxed{ \tt{ \: sin\dfrac{\pi}{4} = \frac{1}{ \sqrt{2} } \: }}$

So, on substituting these values, we get

$\rm = \: \sqrt{2}\bigg[\dfrac{1}{ \sqrt{2} } \: cosx \: - \: \dfrac{1}{ \sqrt{2}} \: sinx\bigg]$

$\rm = \: \sqrt{2}\bigg[\dfrac{cosx - sinx}{ \sqrt{2} } \: \bigg]$

$\rm = \: cosx - sinx$

Hence,

$\rm :\longmapsto\: \boxed{ \tt{ \: \sqrt{2} \: cos\bigg[\dfrac{\pi}{4} + x \bigg] \: = \: cosx - sinx \: }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Explore more :-

Trigonometry Formulas

sin(−θ) = −sin θ

cos(−θ) = cos θ

tan(−θ) = −tan θ

cosec(−θ) = −cosecθ

sec(−θ) = sec θ

cot(−θ) = −cot θ

Product to Sum Formulas

sin x sin y = 1/2 [cos(x–y) − cos(x+y)]

cos x cos y = 1/2[cos(x–y) + cos(x+y)]

sin x cos y = 1/2[sin(x+y) + sin(x−y)]

cos x sin y = 1/2[sin(x+y) – sin(x−y)]

Sum to Product Formulas

sin x + sin y = 2 sin [(x+y)/2] cos [(x-y)/2]

sin x – sin y = 2 cos [(x+y)/2] sin [(x-y)/2]

cos x + cos y = 2 cos [(x+y)/2] cos [(x-y)/2]

cos x – cos y = -2 sin [(x+y)/2] sin [(x-y)/2]

Sum or Difference of angles

cos (A + B) = cos A cos B – sin A sin B

cos (A – B) = cos A cos B + sin A sin B

sin (A+B) = sin A cos B + cos A sin B

sin (A -B) = sin A cos B – cos A sin B

tan(A+B) = [(tan A + tan B)/(1 – tan A tan B)]

tan(A-B) = [(tan A – tan B)/(1 + tan A tan B)]

cot(A+B) = [(cot A cot B − 1)/(cot B + cot A)]

cot(A-B) = [(cot A cot B + 1)/(cot B – cot A)]

cos(A+B) cos(A–B)=cos^2A–sin^2B=cos^2B–sin^2A

sin(A+B) sin(A–B) = sin^2A–sin^2B=cos^2B–cos^2A

Multiple and Submultiple angles

sin2A = 2sinA cosA = [2tan A /(1+tan²A)]

cos2A = cos²A–sin²A = 1–2sin²A = 2cos²A–1= [(1-tan²A)/(1+tan²A)]

tan 2A = (2 tan A)/(1-tan²A)

Answer 2

Answer:

$\large\underline{\sf{Solution-}}$

Consider,

$\rm :\longmapsto\: \sqrt{2} \: cos\bigg[\dfrac{\pi}{4} + x \bigg] \:$

We know,

$\boxed{ \tt{ \: cos(x + y) = cosx \: cosy \: - \: sinx \: siny \: }}$

So, using this

$\rm = \: \sqrt{2}\bigg[cosx \: cos\dfrac{\pi}{4} - sinx \: sin\dfrac{\pi}{4}\bigg]$

We know,

$\: \: \: \: \: \boxed{ \tt{ \: cos\dfrac{\pi}{4} = \frac{1}{ \sqrt{2} } \: }} \: and \: \: \boxed{ \tt{ \: sin\dfrac{\pi}{4} = \frac{1}{ \sqrt{2} } \: }}$

So, on substituting these values, we get

$\rm = \: \sqrt{2}\bigg[\dfrac{1}{ \sqrt{2} } \: cosx \: - \: \dfrac{1}{ \sqrt{2}} \: sinx\bigg]$

$\rm = \: \sqrt{2}\bigg[\dfrac{cosx - sinx}{ \sqrt{2} } \: \bigg]$

$\rm = \: cosx - sinx$

Hence,

$\rm :\longmapsto\: \boxed{ \tt{ \: \sqrt{2} \: cos\bigg[\dfrac{\pi}{4} + x \bigg] \: = \: cosx - sinx \: }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Explore more :-

Trigonometry Formulas

sin(−θ) = −sin θ

cos(−θ) = cos θ

tan(−θ) = −tan θ

cosec(−θ) = −cosecθ

sec(−θ) = sec θ

cot(−θ) = −cot θ

Product to Sum Formulas

sin x sin y = 1/2 [cos(x–y) − cos(x+y)]

cos x cos y = 1/2[cos(x–y) + cos(x+y)]

sin x cos y = 1/2[sin(x+y) + sin(x−y)]

cos x sin y = 1/2[sin(x+y) – sin(x−y)]

Sum to Product Formulas

sin x + sin y = 2 sin [(x+y)/2] cos [(x-y)/2]

sin x – sin y = 2 cos [(x+y)/2] sin [(x-y)/2]

cos x + cos y = 2 cos [(x+y)/2] cos [(x-y)/2]

cos x – cos y = -2 sin [(x+y)/2] sin [(x-y)/2]

Sum or Difference of angles

cos (A + B) = cos A cos B – sin A sin B

cos (A – B) = cos A cos B + sin A sin B

sin (A+B) = sin A cos B + cos A sin B

sin (A -B) = sin A cos B – cos A sin B

tan(A+B) = [(tan A + tan B)/(1 – tan A tan B)]

tan(A-B) = [(tan A – tan B)/(1 + tan A tan B)]

cot(A+B) = [(cot A cot B − 1)/(cot B + cot A)]

cot(A-B) = [(cot A cot B + 1)/(cot B – cot A)]

cos(A+B) cos(A–B)=cos^2A–sin^2B=cos^2B–sin^2A

sin(A+B) sin(A–B) = sin^2A–sin^2B=cos^2B–cos^2A

Multiple and Submultiple angles

sin2A = 2sinA cosA = [2tan A /(1+tan²A)]

cos2A = cos²A–sin²A = 1–2sin²A = 2cos²A–1= [(1-tan²A)/(1+tan²A)]

tan 2A = (2 tan A)/(1-tan²A)