One condition: No calculator!

According to unit circle definition, cosine of an angle is represented on x-axis and sine of an angle is represented on y-axis.

Also,

$\sin135°+\cos135°=0$

Which means that it is the mid point of circle.

Now from the circle it is evident that $|\sin130°|>|\cos130°|$ and $\cos130°$ is negative,

So,

Clearly $\sin130°+\cos130°>0$

We can use the identity $\sin\theta+\cos\theta=\sqrt{2}\sin(\theta+45^\circ)$ :

$R=\sin 130^\circ+\cos 130^\circ=\sqrt{2}\sin(175^\circ)$

Since $\sin\theta>0$ for $0^\circ<\theta<180^\circ$ , then $R>0$ .

$\cos 130^{\circ} = - \cos 50^{\circ} = - \sin 40^{\circ}$ $\sin 130^{\circ} = \sin 50^{\circ}$ $\sin 50^{\circ} > \sin 40^{\circ} \Rightarrow \sin 50^{\circ} - \sin 40^{\circ} > 0$

We rewrite the terms as $\large \sin 50 - \cos 50$ . Now, $\large cos$ trails $\large sin$ in the domain $\large (45,90)$ , where the numbers are expressed in degrees.

Hence, R > 0.

Just look at the graph provided, $130^o$ is slightly less than $\frac{3 \pi }{4}$ . At $\frac{3\pi}{4}$ sum of the graph is 0. Slightly to its left, sum of the graph is > 0.

sin A + cos A is greater than zero if (0<A<135°) U (315°<A≤360°) and their respective coterminals. Since 130° < 135°, hence, R > 0.

Correction: A ≥ 0. Sorry i cannot edit it through my mobile phone.

sin130 + cos130 = sin50 - sin40 = 2sin5 × cos45 = √2 × sin 5 > 0.

This was the only solution that wouldn't have 2 answers

R = sin130° + cos130°

Squaring both sides gives you

R² = sin²130° + cos²130° + 2sin130°.cos130°

From trig identities,

sin²(∅) + cos²(∅) = 1, where ∅ is an angle.

2sin(∅).cos(∅) = sin(2∅), where ∅ is an angle.

By substituting this into the equation we have,

R² = 1 + sin(2×130°)

Further simplification

R² = 1 + sin(260°)

Since the angle 260° is in the third quadrant, we know that sin(260) is negative.

Also, since the angle 260° is greater than 180° and less than 270°, it means that sin(260°) is less than 0 (because sin180° = 0) but greater than -1 (because sin270° = -1)

Therefore,

R² = 1 - 0.xx (xx is any two digit number, we are ballparking here as we aren't allowed to use a calculator.)

R² = 0.xx

R = √(0.xx)

R = 0.xx

Therefore, R > 0.