Solving Trigonometry Problems: Finding sin θ + cos θ

This problem involves finding the value of the sum of sine and cosine of an angle when their difference is given. We are given the equation:

$$ \sin \theta - \cos \theta = \frac{4}{5} $$

We need to find the value of $ \sin \theta + \cos \theta $. Let's denote the value we want to find as $X$.

$$ X = \sin \theta + \cos \theta $$

A useful technique for problems involving sums and differences of sine and cosine is to square the expressions. Let's consider the squares of both expressions:

$$ (\sin \theta - \cos \theta)^2 = \left(\frac{4}{5}\right)^2 $$

Expanding the left side:

$$ \sin^2 \theta - 2 \sin \theta \cos \theta + \cos^2 \theta = \frac{16}{25} $$

Using the fundamental trigonometric identity $ \sin^2 \theta + \cos^2 \theta = 1 $, we can simplify this to:

$$ 1 - 2 \sin \theta \cos \theta = \frac{16}{25} $$

Now let's consider the expression we want to find, $X = \sin \theta + \cos \theta$, and square it:

$$ X^2 = (\sin \theta + \cos \theta)^2 $$

Expanding this:

$$ X^2 = \sin^2 \theta + 2 \sin \theta \cos \theta + \cos^2 \theta $$

Again using the identity $ \sin^2 \theta + \cos^2 \theta = 1 $:

$$ X^2 = 1 + 2 \sin \theta \cos \theta $$

We have two equations now:

1. $ 1 - 2 \sin \theta \cos \theta = \frac{16}{25} $
2. $ X^2 = 1 + 2 \sin \theta \cos \theta $

From equation (1), we can find the value of $ 2 \sin \theta \cos \theta $:

$$ 2 \sin \theta \cos \theta = 1 - \frac{16}{25} = \frac{25 - 16}{25} = \frac{9}{25} $$

Now, substitute this value into equation (2):

$$ X^2 = 1 + \frac{9}{25} $$

$$ X^2 = \frac{25 + 9}{25} = \frac{34}{25} $$

To find $X$, we take the square root of both sides:

$$ X = \sqrt{\frac{34}{25}} $$

$$ X = \frac{\sqrt{34}}{\sqrt{25}} = \frac{\sqrt{34}}{5} $$

So, the value of $ \sin \theta + \cos \theta $ is $ \frac{\sqrt{34}}{5} $. Since the options provide a positive value, we consider the positive root.

Alternative Approach using an Identity

We can also use the identity $(a-b)^2 + (a+b)^2 = 2(a^2+b^2)$.

Let $a = \sin \theta$ and $b = \cos \theta$. Then the identity becomes:

$$ (\sin \theta - \cos \theta)^2 + (\sin \theta + \cos \theta)^2 = 2(\sin^2 \theta + \cos^2 \theta) $$

We know $ \sin \theta - \cos \theta = \frac{4}{5} $ and $ \sin^2 \theta + \cos^2 \theta = 1 $. Let $ (\sin \theta + \cos \theta) = X $. Substituting the known values:

$$ \left(\frac{4}{5}\right)^2 + (X)^2 = 2(1) $$

$$ \frac{16}{25} + X^2 = 2 $$

Now, solve for $X^2$:

$$ X^2 = 2 - \frac{16}{25} $$

$$ X^2 = \frac{50 - 16}{25} $$

$$ X^2 = \frac{34}{25} $$

Taking the square root:

$$ X = \sqrt{\frac{34}{25}} = \frac{\sqrt{34}}{5} $$

Both methods yield the same result.

Summary of Steps to Find sin θ + cos θ

Here is a quick recap of the steps:

1. Start with the given equation: $ \sin \theta - \cos \theta = \frac{4}{5} $.
2. Consider the expression to find: $ \sin \theta + \cos \theta $.
3. Square the given equation: $ (\sin \theta - \cos \theta)^2 = \left(\frac{4}{5}\right)^2 $.
4. Expand and use $ \sin^2 \theta + \cos^2 \theta = 1 $ to find $ 2 \sin \theta \cos \theta $.
5. Square the expression to find: $ (\sin \theta + \cos \theta)^2 $.
6. Expand and use $ \sin^2 \theta + \cos^2 \theta = 1 $ and the value of $ 2 \sin \theta \cos \theta $.
7. Take the square root to find the value of $ \sin \theta + \cos \theta $.

Revision Table: Key Trigonometry Concepts

Additional Information: Understanding sin θ and cos θ Sum/Difference

The sum and difference of sine and cosine, $ \sin \theta + \cos \theta $ and $ \sin \theta - \cos \theta $, are related. Squaring these expressions often simplifies problems because the $ \sin^2 \theta + \cos^2 \theta $ term appears, which is equal to 1.

Consider the general forms:

• $ (\sin \theta + \cos \theta)^2 = \sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = 1 + 2 \sin \theta \cos \theta $
• $ (\sin \theta - \cos \theta)^2 = \sin^2 \theta + \cos^2 \theta - 2 \sin \theta \cos \theta = 1 - 2 \sin \theta \cos \theta $

Adding these two equations gives:

$$ (\sin \theta + \cos \theta)^2 + (\sin \theta - \cos \theta)^2 = (1 + 2 \sin \theta \cos \theta) + (1 - 2 \sin \theta \cos \theta) $$

$$ (\sin \theta + \cos \theta)^2 + (\sin \theta - \cos \theta)^2 = 2 $$

This identity $ (a+b)^2 + (a-b)^2 = 2(a^2+b^2) $ where $a=\sin \theta$ and $b=\cos \theta$ is particularly useful. It allows us to find the value of the sum if the difference is known (or vice versa) without explicitly calculating $ \sin \theta $ or $ \cos \theta $ individually, or the product $ \sin \theta \cos \theta $. This is exactly what was done in the alternative approach.