Question

sin 45°cos 45° - sin 30°​

Answer 1

We know that:

$\sin( {45}^{0} ) = \frac{1}{ \sqrt{2} }$

$\cos( {45}^{0} ) = \frac{1}{ \sqrt{2} }$

$\sin( {30}^{0} ) = \frac{1}{2}$

Now, coming back to your question-

$\sin( {45}^{0} ) . \cos( {45}^{0} ) - \sin( {30}^{0} )$

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

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

$= 0$

Hence, this is the required answer.

Answer 2

Answer:

From geometry,

\sf\angle ACB = \angle BAC = 45∠ACB=∠BAC=45 ,

AB = a , then BC = aFrom Pythagoras theorem ,

\sf (AC)^{2} = (AB)^{2} + (BC)^{2}(AC)

2

=(AB)

2

+(BC)

2

\sf (AC)^{2} = a^{2} + a^{2}(AC)

2

=a

2

+a

2

\sf (AC)^{2} = 2a^{2}(AC)

2

=2a

2

.

Applying square root on both sides we get :-

AC = \sf\sqrt{2a^{2}}

2a

2

= \sqrt{2}a=

2

a

From ∆ABC ,

\sf\angle A = 45∠A=45 ,

we get :-

sin45° = \dfrac{1}{root2}

root2

1

= \frac{a}{ \sqrt{2}a}

2

a

a

= \frac{1}{ \sqrt{2} }

2

1

\sf\therefore sin45 = \dfrac{1}{\sqrt{2}}∴sin45=

2

1