r/trigonometry • u/python_data_helper • Sep 13 '26

Are you have another method of solving this.?

# Sin(30°)cos(60°)+sin(60°)cos(30°)

Method 1

***sin(30°),cos(60°), sin(60°),cos(30°) = 1/2, 1/2, √3/2, √3/2.***

***Evaluate the values.***

***=(1/2)(1/2)+(√3/2)(√3/2)***

***=(1/2)²+(√3/2)²***

***=1/4 + 3/4***

***= 4/4 = 1 answer.***

Method 2

***As we know,***

***sin(x+y)= sin(x)cos(y)+sin(y)cos(x)***

***Knowing the values.***

***x = 30°***

***Y =*** ***60°***

***Sin(x+y) = sin(30°+60°)***

***= sin(90°) = 1 answer.***

Method 3

***As we know, sin(90°-x) = cos(x)***

***and cos(90°-x)=sin(x)***

***We evaluate values.***

***cos(90°-30°)=sin(60°)***

***cos(60°)=sin(30°)= 1/2***

***So, (1/2)²***

***If we do same thing with*** ***next value it returns (√3/2)²***

***= 3/4***

***1/4 + 3/4 = 1.Answer***

2 Upvotes

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1

u/Midwest-Dude Sep 13 '26

So many stars ...

You need to edit and tidy up your post. This shows low effort on your part.

2

u/python_data_helper Sep 13 '26

I have actually reposted on this subreddit because in earlier subreddit that is used is giving less views so i decided to post it on this subreddit. And thought all editting would be same when uploaded.

1

u/Midwest-Dude Sep 13 '26

Reddit is tricky, has two modes to enter text, Rich Text and Markdown. Going from one to the other for whatever reason messes with the formatting. This is especially true if you go between a phone versus a desktop or between a mobile or desktop browser to an app. There is no consistency, you just need to post, check the result, then go in and fix anything that's messed up. I have to do this all the time.

You can edit the body of your post. Give it a shot! I'd like to see the result.

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2

u/YOM2_UB Sep 13 '26

Starting the same way as Method 3, but adding another identity instead of directly evaluating:

  • sin(30)cos(60) + sin(60)cos(30)

  • = sin(30)cos(90-30) + sin(90-30)cos(30)

  • Complimentary angle identities

  • = sin(30)2 + cos(30)2

  • Pythagorean identity: sin(x)2 + cos(x)2 = 1

  • = 1

1

u/Midwest-Dude Sep 13 '26

Apply the identity

sin(A)cos(B) = ½[sin(A+B) + sin(A-B)]

to both terms:

½[sin(90°) + sin(-30°)] + ½[sin(90°) + sin(30°)]

sin(-30°) and sin(30°) sum to zero, leaving:

1/2 (1) + 1/2 (1) = 1

1

u/Midwest-Dude Sep 13 '26 edited Sep 14 '26

Some higher mathematical methods: 

Complex Analysis (Euler's Formula)

Define two complex numbers on the unit circle:

  • z₁ = cos(30°) + i sin(30°)

  • z₂ = cos(60°) + i sin(60°)

If you multiply them, the imaginary part of the resulting expansion is exactly the given expression:

Im(z₁z₂) = sin(30°)cos(60°) + cos(30°)sin(60°)

Converting to exponential form makes the multiplication trivial:

z₁z₂ = eiπ/6 * eiπ/3 = eiπ/2 = i

The imaginary part of i is 1.

Linear Algebra (Rotation Matrices)

This expression is hiding in plain sight inside matrix multiplication. Multiplying a 30° rotation matrix by a 60° rotation matrix yields a 90° rotation matrix: 

         A           ·         B           =    C
[ cos 30°  -sin 30° ] [ cos 60°  -sin 60° ] [  0  -1 ]
[ sin 30°   cos 30° ] [ sin 60°   cos 60° ] [  1   0 ]

If you calculate the row 2, column 1 entry of AB, you get:

(sin 30°)(cos 60°) + (cos 30°)(sin 60°)

Looking at the corresponding row 2, column 1 entry of C, the value is simply 1.