Question

if tanΘ= root 2-1 , prove that sonΘcosΘ= root2 /4

Answer 1

Step-by-step explanation:

As per the information given in the question, We have :

• Tan θ = √2 - 1

We are asked to prove that sin θ cos θ = √2/4

Here, In this question we have to find sin θ & cos θ first, Then we will put those values in sin θ cos θ & Simplify it until it becomes √2/4.

$\longmapsto \rm \tan \theta = \dfrac{opp}{adj}$

$\longmapsto \rm \tan \theta = \dfrac{ \sqrt{2} - 1}{1}$

« Now, By using Pythagoras theorm,

⌬ (AC)² = (AB)² + (BC)²

→ (hypotenuse)² = (opposite)² + (adjacent)²

→ (h)² = (√2 - 1)² + (1)²

→ (h)² = 3 - 2√2 + 1

→ (h)² = 4 - 2√2

→ h = √4 - 2√2

$\longmapsto \rm \sin \theta = \dfrac{opp}{hyp}$

$\longmapsto \rm \sin \theta = \dfrac{ \sqrt{2} - 1}{ \sqrt{4 - 2\sqrt{2} } }$

$\longmapsto \rm \cos \theta = \dfrac{adj}{hyp }$

$\longmapsto \rm \cos \theta = \dfrac{1}{ \sqrt{4 - 2 \sqrt{2}} }$

$\longmapsto \rm \sin \theta \cos \theta = \dfrac{1}{ \sqrt{4 - 2 \sqrt{2}} } \times \dfrac{ \sqrt{2} - 1}{ \sqrt{4 - 2\sqrt{2}} }$

∴ ⌬ √a √a = a

$\longmapsto \rm \sin \theta \cos \theta = \dfrac{\sqrt{2} - 1}{ 4 - 2 \sqrt{2} }$

On rationalizing the denominator,

$\longmapsto \rm \sin \theta \cos \theta = \dfrac{\sqrt{2} - 1}{ 4 - 2 \sqrt{2} } \times \dfrac{4 + 2 \sqrt{2} }{4 + 2 \sqrt{2} }$

$\longmapsto \rm \sin \theta \cos \theta = \dfrac{(\sqrt{2} - 1)(4 + 2 \sqrt{2} )}{ (4 - 2 \sqrt{2})(4 + 2 \sqrt{2}) }$

$\longmapsto \rm \sin \theta \cos \theta = \dfrac{4 \sqrt{2} - 4 + 4 - 2 \sqrt{2} }{ {(4)}^{2} - {(2 \sqrt{2} )}^{2} }$

$\longmapsto \rm \sin \theta \cos \theta = \dfrac{4 \sqrt{2} - 2 \sqrt{2} }{ 16 - 8 }$

$\longmapsto \rm \sin \theta \cos \theta = \dfrac{2 \sqrt{2} }{ 8}$

$\longmapsto \rm \sin \theta \cos \theta = \dfrac{ \sqrt{2} }{ 4}$

Hence, proved!

⠀

M o r e⠀F o r m u l a e :

⠀

• $\sf sin ( A + B ) = \bf{sin A cos B + cos A sin B}$

• $\sf cos ( A + B ) = \bf{cos A cos B - sin A sin B}$

• $\sf tan ( A + B ) = \bf{tan A + tan B/1 - tan A tan B}$

• $\sf tan ( A - B ) = \bf{ tan A - tan B/1 + tan A tan B}$

• $\sf cos ( A - B ) = \bf{ cos A cos B - sin A sin B}$

• $\sf sin ( A - B ) = \bf{ sin A cos B - cos A sin B}$

Answer 2

Question :

if tan θ = √2-1 , prove that sin θ cos θ= √2 /

Solution :

So according to the problem tan θ = √2 – 1

tan θ = Perpendicular/Base

√2 - 1 can be written as :

$\frac{ \sqrt{2} - 1 }{1}$

So, Base = √2 – 1 and the perpendicular = 1

So as in the pic we can see the hypotenuse is not given so we need to apply trigonometry to find out the hypotenuse

By applying pythagoras theorem we get

(Hypotentuse)² = (Perpendicular)² + (Base)²

(Hypotentuse)² = (√2 – 1)² + (1)²

(Hypotentuse)² = (√2)² – 2•√2•1 + (1)² + (1)²

(Hypotentuse)² = 2 – 2√2 + 1 + 1

(Hypotentuse)² = 2 – 2√2 + 2

(Hypotentuse)² = 4 – 2√2

$\sf Hypotentuse = \sqrt{4 - 2 \sqrt{2} }$

Using sin θ formula we get

sin θ = Perpendicular/Hypotentuse

$\sf sin \: θ = \frac{ \sqrt{2} - 1}{ \sqrt{4 - 2 \sqrt{2} } }$

cos θ = Base/Hypotentuse

$\sf cos \: \theta = \frac{1}{ \sqrt{4-2\sqrt{2}}}$

Now as given we need to prove

sin θ cos θ = √2/4

By putting the values we get

$\sf \frac{ \sqrt{2} - 1}{ \sqrt{4 - 2 \sqrt{2} } } \times \frac{1}{ \sqrt{4 - 2 \sqrt{2} } }$

The denominator is common so it will be squared

$\sf \frac{ \sqrt{2} - 1}{ {( \sqrt{4 - 2 \sqrt{2} } })^{2} } = \frac{ \sqrt{2} }{4}$

$\sf\frac{ \sqrt{2} - 1}{4 - 2 \sqrt{2} } = \frac{ \sqrt{2} }{4}$

Now by rationalising the denominator on the L.H.S. we get

$\sf \frac{( \sqrt{2} - 1)(4 + 2 \sqrt{2}) }{(4 - 2 \sqrt{2} )(4 + 2 \sqrt{2}) }$

$\sf\frac{4 \sqrt{2} + 4 - 4 - 2 \sqrt{2} }{ {(4)}^{2} - {(2 \sqrt{2}) }^{2} } = \frac{ \sqrt{2} }{4}$

$\sf \frac{4 \sqrt{2} + \cancel{ 4} - \cancel{4} - 2 \sqrt{2} }{16 - 8} = \frac{ \sqrt{2} }{4}$

$\sf\frac{2 \sqrt{2} }{8} = \frac{ \sqrt{2} }{4}$

$\sf \frac{ \cancel{2} \sqrt{2} }{ \cancel{8}} = \frac{ \sqrt{2} }{4}$

$\sf \frac{ \sqrt{2} }{4} = \frac{ \sqrt{2} }{4}$

Hence, Proved !!