The value of sin 70 degrees + cos 40 Degrees by cos 70 degrees plus sin 40 Degrees

The value of sin 70 degrees + cos 40 Degrees by cos 70 degrees plus sin 40 Degrees

The Correct Answer and Explanation is:

To evaluate the expression: sin⁡70∘+cos⁡40∘bycos⁡70∘+sin⁡40∘\sin 70^\circ + \cos 40^\circ \quad \text{by} \quad \cos 70^\circ + \sin 40^\circsin70∘+cos40∘bycos70∘+sin40∘

First, recall that: cos⁡(90∘−x)=sin⁡(x)andsin⁡(90∘−x)=cos⁡(x)\cos(90^\circ – x) = \sin(x) \quad \text{and} \quad \sin(90^\circ – x) = \cos(x)cos(90∘−x)=sin(x)andsin(90∘−x)=cos(x)

Step 1: Express terms using co-function identities

We know that:

• cos⁡40∘=sin⁡(90∘−40∘)=sin⁡50∘\cos 40^\circ = \sin (90^\circ – 40^\circ) = \sin 50^\circcos40∘=sin(90∘−40∘)=sin50∘
• sin⁡40∘=cos⁡(90∘−40∘)=cos⁡50∘\sin 40^\circ = \cos (90^\circ – 40^\circ) = \cos 50^\circsin40∘=cos(90∘−40∘)=cos50∘

Now, we can rewrite the expression as: sin⁡70∘+cos⁡40∘becomessin⁡70∘+sin⁡50∘\sin 70^\circ + \cos 40^\circ \quad \text{becomes} \quad \sin 70^\circ + \sin 50^\circsin70∘+cos40∘becomessin70∘+sin50∘ cos⁡70∘+sin⁡40∘becomescos⁡70∘+cos⁡50∘\cos 70^\circ + \sin 40^\circ \quad \text{becomes} \quad \cos 70^\circ + \cos 50^\circcos70∘+sin40∘becomescos70∘+cos50∘

Step 2: Simplify using sum of angles formula

We can use the sine and cosine addition formulas for further simplification.

• The sum of sine terms:

sin⁡A+sin⁡B=2sin⁡(A+B2)cos⁡(A−B2)\sin A + \sin B = 2 \sin\left(\frac{A + B}{2}\right) \cos\left(\frac{A – B}{2}\right)sinA+sinB=2sin(2A+B​)cos(2A−B​)

• The sum of cosine terms:

cos⁡A+cos⁡B=2cos⁡(A+B2)cos⁡(A−B2)\cos A + \cos B = 2 \cos\left(\frac{A + B}{2}\right) \cos\left(\frac{A – B}{2}\right)cosA+cosB=2cos(2A+B​)cos(2A−B​)

For sin⁡70∘+sin⁡50∘\sin 70^\circ + \sin 50^\circsin70∘+sin50∘, applying the sine addition formula: sin⁡70∘+sin⁡50∘=2sin⁡(70∘+50∘2)cos⁡(70∘−50∘2)\sin 70^\circ + \sin 50^\circ = 2 \sin\left(\frac{70^\circ + 50^\circ}{2}\right) \cos\left(\frac{70^\circ – 50^\circ}{2}\right)sin70∘+sin50∘=2sin(270∘+50∘​)cos(270∘−50∘​) =2sin⁡60∘cos⁡10∘= 2 \sin 60^\circ \cos 10^\circ=2sin60∘cos10∘

For cos⁡70∘+cos⁡50∘\cos 70^\circ + \cos 50^\circcos70∘+cos50∘, applying the cosine addition formula: cos⁡70∘+cos⁡50∘=2cos⁡(70∘+50∘2)cos⁡(70∘−50∘2)\cos 70^\circ + \cos 50^\circ = 2 \cos\left(\frac{70^\circ + 50^\circ}{2}\right) \cos\left(\frac{70^\circ – 50^\circ}{2}\right)cos70∘+cos50∘=2cos(270∘+50∘​)cos(270∘−50∘​) =2cos⁡60∘cos⁡10∘= 2 \cos 60^\circ \cos 10^\circ=2cos60∘cos10∘

Step 3: Final simplification

We know: sin⁡60∘=32andcos⁡60∘=12\sin 60^\circ = \frac{\sqrt{3}}{2} \quad \text{and} \quad \cos 60^\circ = \frac{1}{2}sin60∘=23​​andcos60∘=21​

Thus, the expression becomes: 2sin⁡60∘cos⁡10∘=2×32×cos⁡10∘=3cos⁡10∘2 \sin 60^\circ \cos 10^\circ = 2 \times \frac{\sqrt{3}}{2} \times \cos 10^\circ = \sqrt{3} \cos 10^\circ2sin60∘cos10∘=2×23​​×cos10∘=3​cos10∘ 2cos⁡60∘cos⁡10∘=2×12×cos⁡10∘=cos⁡10∘2 \cos 60^\circ \cos 10^\circ = 2 \times \frac{1}{2} \times \cos 10^\circ = \cos 10^\circ2cos60∘cos10∘=2×21​×cos10∘=cos10∘

Step 4: Final result

So, the original expression simplifies to: 3cos⁡10∘cos⁡10∘=3\frac{\sqrt{3} \cos 10^\circ}{\cos 10^\circ} = \sqrt{3}cos10∘3​cos10∘​=3​

Thus, the value of sin⁡70∘+cos⁡40∘\sin 70^\circ + \cos 40^\circsin70∘+cos40∘ by cos⁡70∘+sin⁡40∘\cos 70^\circ + \sin 40^\circcos70∘+sin40∘ is approximately: 3\boxed{\sqrt{3}}3​​